12th NCERT INTEGRALS Exercise 7.2 Questions 39
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Integrate the functions in Exercise 1 to 37

Question (1)

\[\frac{{2x}}{{1 + {x^2}}}\] Solution: \[I = \int {\frac{{2x}}{{1 + {x^2}}}dx} \]
Because of x in numerator we can not use the formula of \[\int {\frac{1}{{{x^2} + {a^2}}}} dx\]
Let 1+x2 = t
Differentiate w.r.t. x 2xdx = dt
Replacing value of 2xdx and 1+x2 we get \[\begin{array}{l}I = \int {\frac{{2x}}{{1 + {x^2}}}dx} \\I = \int {\frac{{dt}}{t}} \end{array}\]
Formula \[\int {\frac{1}{x}dx = \log \left| x \right|} + c\]
I = log|t|+c
replacing value of t in above equation
I=log|1+x2|+c

Question (2)

\[\frac{{{{\left( {\log x} \right)}^2}}}{x}\] Solution: \[I = \int {\frac{{{{\left( {\log x} \right)}^2}}}{x}dx} \] Normally composite function is assumed as t. So here we will assume log x = t. Let logx = t
dx/x = dt
\[\begin{array}{l}I = {\int {\left( {\log x} \right)} ^2} \cdot \frac{1}{x}dx\\I = \int {{t^2}} dt\\I = \frac{{{t^3}}}{3} + c\end{array}\] Substituting value of t
\[I = \frac{{{{\left( {\log x} \right)}^3}}}{3} + c\]

Question (3)

\[\frac{1}{{x + x\log x}}\] Solution: \[\begin{array}{l}I = \int {\frac{1}{{x + x\log x}}} dx\\I = \int {\frac{1}{{x\left( {1 + \log x} \right)}}dx} \end{array}\] Let 1+logx = t
Differtiate w.r.t. x
dx/x = dt
\[\begin{array}{l}I = \int {\frac{1}{{\left( {1 + \log x} \right)}} \cdot \frac{1}{x}dx} \\I = \int {\frac{1}{t}dt} \\I = \log |t| + c\end{array}\] \[I = \log \left| {1 + \log x} \right| + c\]

Question (4)

sinx sin(cosx)
Solution: \[I = \int {\sin x\sin \left( {\cos x} \right)} dx\]
As there is no formula for multiplication with same function. Solve substitution.
Let cosx = t
∴ -sinxdx = dt
sinxdx = -dt
\[I = \int {\sin \left( {\cos x} \right)\sin x} dx\] Substituting above values in equation
\[\begin{array}{l}I = \int {\sin \left( t \right)} \left( { - dt} \right)\\I = - \int {\sin tdt} \\I = - \left( { - \cos t} \right) + c\\I = \cos t + c\\I = \cos \left( {\cos x} \right) + c\end{array}\]

Question (5)

sin(ax+b)cos(ax+b)
Solution:\[I = \int {\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx} \]
Multiply by 2 so that we can use formula of 2sinθcosθ = sin2θ
\[\begin{array}{l}I = \frac{1}{2}\int {2\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx} \\I = \frac{1}{2}\int {\sin 2\left( {ax + b} \right)dx} \end{array}\]
Use formula \[\int {\sin a\theta = - \frac{{\cos \theta }}{a}} + c\]
\[\begin{array}{l}I = \frac{1}{2}\frac{{ - \cos 2\left( {ax + b} \right)}}{{2a}} + c\\I = \frac{{ - \cos 2\left( {ax + b} \right)}}{{4a}} + c\end{array}\] OTHER METHOD
Let sin(ax+b) = t
∴ cos(ax+b) . a dx = dt
∴ cos(ax+b) dx = dt/a
\[\begin{array}{l}I = \int {\sin \left( {ax + b} \right) \cdot \cos \left( {ax + b} \right)} dx\\I = \int {t\frac{{dt}}{a}} \\I = \frac{1}{a}\int {tdt} \\I = \frac{{{t^2}}}{{2a}} + c'\end{array}\] On substituting value of "t"
\[\begin{array}{l}I = \frac{{{t^2}}}{{2a}} + c'\\I = \frac{{{{\sin }^2}\left( {ax + b} \right)}}{{2a}} + c'\end{array}\]
Using trigonometric identities \[{\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}\]
\[\begin{array}{l}I = \frac{{{t^2}}}{{2a}} + c'\\I = \frac{1}{{2a}}\left[ {\frac{{1 - \cos 2\left( {ax + b} \right)}}{2}} \right] + c'\\I = \frac{1}{{4a}} - \frac{1}{{4a}}\cos 2\left( {ax + b} \right) + c'\\let\left[ {c = c' + \frac{1}{{4a}}} \right]\\I = - \frac{1}{{4a}}\cos 2\left( {ax + b} \right) + c\end{array}\]

Question (6)

\[\sqrt {ax + b} \] Solution: \[I = \int {\sqrt {ax + b} }\quad dx \] We will assume ax + b = t2 , so that it will simplify in t we can avoide square root sign.
Let ax+b=t2
Differentiate w.r.t. x
\[\begin{array}{l}a = 2t\frac{{dt}}{{dx}}\\dx = \frac{{2t}}{a}dt\end{array}\] Substituting values in given equation \[\begin{array}{l}I = \int {\sqrt {{t^2}} } \cdot \frac{{2t}}{a}dt\\I = \frac{2}{a}\int {t \cdot tdt} \\I = \frac{2}{a}\int {{t^2} \cdot dt} \\I = \frac{2}{a}\frac{{{t^3}}}{3} + c\\I = \frac{2}{{3a}}{\left[ {{{\left( {ax + b} \right)}^{\frac{1}{2}}}} \right]^3} + c\end{array}\] \[I = \frac{2}{{3a}}{\left( {ax + b} \right)^{\frac{3}{2}}} + c\]

Question (7)

\[x\sqrt {x + 2} \] Solution: \[I = \int {x\sqrt {x + 2} \cdot d} x\] here we can not multiply directly
Let x+2 = t2
and t = √(x+2)
x = t2 - 2
∴ dx = 2t dt
Substituting values in equation \[\begin{array}{l}I = \int {\left( {{t^2} - 2} \right)\sqrt {{t^2}} \cdot 2tdt} \\I = \int {\left( {{t^2} - 2} \right)t} \cdot 2tdt\\I = \int {2{t^2}} \left( {{t^2} - 2} \right)dt\\I = 2\int {{t^4}dt - 4\int {{t^2}dt} } \\I = \frac{2}{5}{t^5} - \frac{4}{3}{t^3} + c\end{array}\] Substituting value of "t" \[I = \frac{2}{5}{\left( {x + 2} \right)^{\frac{5}{2}}} - \frac{4}{5}{\left( {x + 2} \right)^{\frac{3}{2}}} + c\]

Question (8)

\[x\sqrt {1 + 2{x^2}} \] Solution: \[I = \int {x\sqrt {1 + 2{x^2}} \cdot dx} \] Let 1+2x2 = t2
∴ t = (1+2x2)1/2
∴ 4x dx = 2t dt
x dx = (t/2) dt
Substituting values in given equation \[\begin{array}{l}I = \int {\sqrt {{t^2}} } .\frac{t}{2}dt\\I = \frac{1}{2}\int {{t^2}dt} \\I = \frac{1}{2}\frac{{{t^3}}}{3} + c\\I = \frac{{{t^3}}}{6} + c\\I = \frac{1}{6}{\left( {1 + 2{x^2}} \right)^{\frac{3}{2}}} + c\end{array}\]

Question (9)

\[\left( {4x + 2} \right)\sqrt {{x^2} + x + 1} \] Solution: \[\begin{array}{l}I = \int {\left( {4x + 2} \right)\sqrt {{x^2} + x + 1} \;} dx\\I = 2\int {\left( {2x + 1} \right)} \sqrt {{x^2} + x + 1} \;dx\end{array}\] Assume composite function as t2
∴ x2 + x + 1 = t2
Taking diff. w.r.t x
(2x+1) dx = 2t dt
t= (x2 + x + 1)1/2
Substituting values in given equation \[\begin{array}{l}I = 2\int {\left( {2x + 1} \right)} \sqrt {{x^2} + x + 1} \;dx\\I = 2\int {\sqrt {{t^2}} } \cdot 2tdt\\I = 4\int {t \cdot tdt} \\I = 4\int {{t^2}dt} \\I = \frac{4}{3}{t^3}+c\\I = \frac{4}{3}{\left( {{x^2} + x + 1} \right)^{\frac{3}{2}}} + c\end{array}\]

Question (10)

\[\frac{1}{{x - \sqrt x }}\] Solution: \[I = \int {\frac{1}{{x - \sqrt x }}dx} \] Let x =t2
∴ t = √x
∴ dx = 2t dt
\[\begin{array}{l}I = \int {\frac{1}{{{t^2} - \sqrt {{t^2}} }} \cdot 2tdt} \\I = 2\int {\frac{t}{{{t^2} - t}}} dt\\I = 2\int {\frac{t}{{t\left( {t - 1} \right)}}} dt\\I = 2\int {\frac{1}{{\left( {t - 1} \right)}}} dt\\I = 2\log \left| {t - 1} \right| + c\\I = 2\log \left| {\sqrt x - 1} \right| + c\end{array}\]

Question (11)

\[\frac{x}{{\sqrt {x + 4} }},x > 0\] Solution: \[I = \int {\frac{x}{{\sqrt {x + 4} }},x > 0} \] Let x + 4 = t2
t=(x+4)1/2
x = t2 - 4
dx = 2tdt
\[\begin{array}{l}I = \int {\frac{{\left( {{t^2} - 4} \right)}}{{\sqrt {{t^2}} }} \cdot 2tdt} \\I = \int {\frac{{\left( {{t^2} - 4} \right)}}{t}} \cdot 2tdt\\I = \int {2\left( {{t^2} - 4} \right)dt} \\I = \int {2{t^2}dt - \int {8dt} } \\I = 2\frac{{{t^3}}}{3} - 8t + c\\I = \frac{2}{3}t\left[ {{t^2} - 12} \right] + c\end{array}\] On substituting value of t and t2 in above \[\begin{array}{l}I = \frac{2}{3}{\left( {x + 4} \right)^{\frac{1}{2}}}\left( {x + 4 - 12} \right) + c\\I = \frac{2}{3}{\left( {x + 4} \right)^{\frac{1}{2}}}\left( {x - 8} \right) + c\end{array}\]

Question (12)

\[{\left( {{x^3} - 1} \right)^{\frac{1}{3}}}{x^5}\] Solution \[I = \int {{{\left( {{x^3} - 1} \right)}^{\frac{1}{3}}}{x^5}}dx \] Let x3 -1 = t3
By diff. above equation w.r.t "t"
∴ 3x2dx = 3t2dt
or x2dx = t2dt
\[I = \int {{{\left( {{x^3} - 1} \right)}^{\frac{1}{3}}}{x^5}} dx\] \[\begin{array}{l}I = \int {{{\left( {{x^3} - 1} \right)}^{\frac{1}{3}}}{x^5}} dx\\I = \int {{{\left( {{x^3} - 1} \right)}^{\frac{1}{3}}}{x^3}{x^2}} dx\end{array}\] Substituting values of t and x3, x2dx in above equation we get
\[\begin{array}{l}I = \int {{{\left( {{t^3}} \right)}^{\frac{1}{3}}}} \left( {{t^3} + 1} \right){t^2}dt\\I = \int t \left( {{t^3} + 1} \right){t^2}dt\\I = \int {{t^6}} dt + \int {{t^3}dt} \\I = \frac{1}{7}{t^7} + \frac{1}{4}{t^4} + c\end{array}\] By substituting value of "t" in above equation
\[I = \frac{1}{7}{\left( {{x^3} - 1} \right)^{\frac{7}{3}}} + \frac{1}{4}{\left( {{x^3} - 1} \right)^{\frac{4}{3}}} + c\]

Question (13)

\[\frac{{{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}\] Solution: \[I = \int {\frac{{{x^2}}}{{{{\left( {2 + 3{x^3}} \right)}^3}}}dx} \] Let 2+3x3 = t
Taking differentiation \[{{x^2}dx = \frac{{dt}}{9}}\] Substituting values in given equation \[{\begin{array}{*{20}{l}}{I = \int {\frac{1}{{9{t^3}}}} dt}\\{I = \frac{1}{9}\int {{t^{ - 3}}dt} }\\{I = \frac{1}{9}\frac{{{t^{ - 2}}}}{{ - 2}} + c}\\{}\end{array}}\] Substituting value of "t" \[I = \frac{{ - 1}}{{18{{\left( {2 + 3{x^3}} \right)}^2}}}+c\]

Question (14)

\[\frac{1}{{x{{\left( {\log x} \right)}^m}}},x > 0\] Solution \[I = \int {\frac{1}{{x{{\left( {\log x} \right)}^m}}}dx} ,x > 0\] Let t = logx Taking diff. dt = dx/x
Substituting values in given equation
\[\begin{array}{l}I = \int {\frac{{dt}}{{{t^m}}}} \\I = \frac{{{t^{ - m + 1}}}}{{ - m + 1}} + c\end{array}\] Substituting value of "t"
\[I = \frac{{{{\left( {\log x} \right)}^{1 - m}}}}{{1 - m}} + c\]

Question (15)

\[\frac{x}{{9 - 4{x^2}}}\] Solution: \[I = \int {\frac{x}{{9 - 4{x^2}}}dx} \] Let 9-4x2 = t
Taking diff.t
-8xdx = dt
∴ xdx= -dt/8
Substituting values in given equation \[\begin{array}{l}I = \int {\frac{{\frac{{ - dt}}{8}}}{t}} \\I = - \frac{1}{8}\int {\frac{1}{t}} dt\\I = - \frac{1}{8}\log t + c\\substitute\ value\ of\ t\\I = - \frac{1}{8}\log \left( {9 - 4{x^2}} \right) + c\end{array}\]

Question (16)

\[{e^{2x + 3}}\]

Solution:

Let 2x+3 = t
∴ 2dx = dt
dx = dt /2 \[\begin{array}{l}I = \int {{e^{2x + 3}}dx} \\I = \int {{e^t}} \frac{{dt}}{2}\\I = \frac{1}{2}{e^t} + c\\I = \frac{1}{2}{e^{2x + 3}} + c\end{array}\]

Question (17)

\[\frac{x}{{{e^{{x^2}}}}}\]

Solution:

Let x2 = t
2xdx= dt
xdx = dt/2
\[\int {{e^{ - x}}dx = } \frac{{{e^{ - x}}}}{{\frac{d}{{dx}}\left( { - x} \right)}} = - {e^{ - x}}\]
\[\begin{array}{l}I = \int {\frac{{\frac{{dt}}{2}}}{{{e^t}}}} \\I = \frac{1}{2}\int {{e^{ - t}}} dt\\I = \frac{1}{2} \cdot \frac{{{e^{ - t}}}}{{ - 1}} + c\\I = - \frac{1}{{2{e^t}}} + c\\I = - \frac{1}{{2{e^{{x^2}}}}} + c\end{array}\]

Question (18)

\[\frac{{{e^{{{\tan }^{ - 1}}}}x}}{{1 + {x^2}}}dx\]

solution

Let tan-1x = t
Derivative of tan-1x is
\[\frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}}\]
\[{\mkern 1mu} \frac{1}{{1 + {x^2}}}dx = dt\] \[\begin{array}{l}I = \int {{e^{{{\tan }^{ - 1}}x}} \cdot \frac{1}{{1 + {x^2}}}dx} \\I = \int {{e^t}dt} \\I = {e^t} + c\\I = {e^{{{\tan }^{ - 1}}}}x + c\end{array}\]

Question (19)

\[\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}\]

solution

\[I = \int {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}dx} \] Divide numerator and denominator by ex \[I = \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} \] Let ex+e-x = t
∴ [ex+e-x(-1)]dx = dt
∴ (ex-e-x(-1))dx = dt
\[\begin{array}{l}I = \int {\frac{{dt}}{t}} \\I = \log \left| t \right| + c\\I = \log \left| {{e^x} + {e^{ - x}}} \right| + c\end{array}\]

Question (20)

\[\frac{{{e^{2x}} - {e^{ - 2x}}}}{{{e^{2x}} + {e^{ - 2x}}}}\]

solution

\[I = \int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{{{e^{2x}} + {e^{ - 2x}}}}} \] Let e2x+e-2x=t
[e2x.(2) +e-2x.(-2)]dx=dt \[{e^{2x}} - {e^{ - 2x}}dx = \frac{{dt}}{2}\] Substituting above values in given equation \[\begin{array}{l}I = \int {\frac{{\frac{{dt}}{2}}}{t}} \\I = \frac{1}{2}\int {\frac{{dt}}{t}} \\I = \frac{1}{2}\log \left| t \right| + c\\I = \frac{1}{2}\log \left| {{e^{2x}} + {e^{ - 2x}}} \right| + c\end{array}\]

Question (21)

tan2(2x-3)

solution

\[I = \int {{{\tan }^{ 2}}\left( {2x - 3} \right)dx} \]
There is no formula for ∫tan2 so we will use identity tan2 = sec2 - 1
\[\begin{array}{l}I = \int {\left[ {{{\sec }^2}\left( {2x - 3} \right) - 1} \right]} dx\\I = \int {{{\sec }^2}\left( {2x - 3} \right)dx - \int {dx} } \end{array}\] Let 2x-3=t
2dx=dt
dx = dt/2 Substituting above values in given equation \[\begin{array}{l}I = \int {{{\sec }^2}t \cdot \frac{{dt}}{2} - x + c} \\I = \frac{1}{2}\int {{{\sec }^2}tdt - x + c} \\I = \frac{1}{2}\tan t - x + c\\I = \frac{1}{2}\tan \left( {2x - 3} \right) - x + c\end{array}\]

Question (22)

sec2(7-4x)dx

solution

\[I = \int {{{\sec }^2}\left( {7 - 4x} \right)dx} \] Let 7-4x=t
-4dx=dt
dx=dt/-4
Substituting above values in given given \[\begin{array}{l}I = \int {{{\sec }^2}t \cdot \frac{{dt}}{{ - 4}}} \\I = - \frac{1}{4}\int {{{\sec }^2}tdt} \\I = - \frac{1}{4}\tan t + c\end{array}\] Substituting values of "t" in above equation \[I = - \frac{1}{4}\tan \left( {7 - 4x} \right) + c\]

Question (23)

`\[\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\]

solution

\[I = \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx} \] \[I = \int {{{\sin }^{ - 1}}x \cdot \frac{1}{{\sqrt {1 - {x^2}} }}dx} \] Let sin-1x = t
Differtiating w.r.t "x" we get \[\frac{1}{{\sqrt {1 - {x^2}} }}dx = dt\] Substituting values in given equation
\[\begin{array}{l}I = \int {t \cdot dt} \\I = \frac{{{t^2}}}{2} + c\end{array}\] Substituting value of "t" \[I = \frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{2} + c\]

Question (24)

\[\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}\]

solution

\[\begin{array}{l}I = \int {\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}dx} \\I = \frac{1}{2}\int {\frac{{2\cos x - 3\sin x}}{{3\cos x + 2\sin x}}dx} \end{array}\] Let 3cosx + 2sinx = t
∴ [3(-sinx)+2cosx]dx = dt
(2cosx - 3sinx)dx = dt
Substituting above values in given equation \[\begin{array}{l}I = \frac{1}{2}\int {\frac{{dt}}{t}} \\I = \frac{1}{2}\log \left| t \right| + c\\I = \frac{1}{2}\log \left| {3\cos x + 2\sin x} \right| + c\end{array}\]

Question (25)

\[\frac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}}\]

solution

\[\begin{array}{l}I = \int {\frac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}}dx} \\I = \int {\frac{{{{\sec }^2}x}}{{{{\left( {1 - \tan x} \right)}^2}}}} dx\end{array}\] Let 1-tanx = t
∴ -sec2dx = dt
sec2dx = -dt
Substituting above values in given equation \[\begin{array}{l}I = \int {\frac{{ - dt}}{{{t^2}}}} \\I = - \int {{t^{ - 2}}dt} \\I = - \left[ {\frac{{{t^{ - 1}}}}{{ - 1}}} \right] + c\\I = \frac{1}{t} + c\end{array}\] Substituting value of "t" \[I = \frac{1}{{1 - \tan x}} + c\]

Question (26)

\[\frac{{\cos \sqrt x }}{{\sqrt x }}\]

solution

\[I = \int {\frac{{\cos \sqrt x }}{{\sqrt x }}} \] \[I = \int {\cos \sqrt x \cdot \frac{1}{{\sqrt x }}dt} \] Let √x = t
Taking darivative wrt "x" \[\begin{array}{l}\frac{1}{{2\sqrt x }}dx = dt\\\frac{1}{{\sqrt x }}dx = 2dt\end{array}\] Substituting values in given equation \[\begin{array}{l}I = \int {\cos t \cdot 2dt} \\I = 2\int {\cos tdt} \\I = 2\sin t + c\end{array}\] Substituting value of "t" \[I = 2\sin \sqrt x + c\]

Question (27)

\[\sqrt {\sin 2x} \cos 2x\]

solution:

\[I = \int {\sqrt {\sin 2x} \cos 2xdx} \] Let sin2x=t2 ⇒ t=(sin2x)1/2
∴ cos2x.2dx=2t dt
cos2x dx = t dt Substituting values in given equation \[\begin{array}{l}I = \int {\sqrt {{t^2}} \cdot tdt} \\I = \int {t \cdot tdt} \\I = \int {{t^2}dt} \\I = \frac{{{t^3}}}{3} + c\\I = \frac{{{{\left( {\sin 2x} \right)}^{\frac{3}{2}}}}}{3} + c\end{array}\]

Question (28)

\[\frac{{\cos x}}{{\sqrt {1 + \sin x} }}\]

solution:

\[I = \int {\frac{{\cos x}}{{\sqrt {1 + \sin x} }}dx} \] Let 1+sinx = t2 ⇒ t = (1+sinx)1/2
cosx dx = 2tdt
Substituting value in given equation \[\begin{array}{l}I = \int {\frac{{2t}}{{\sqrt {{t^2}} }}dt} \\I = 2\int {\frac{t}{t}dt} \\I = 2t + c\\I = 2{\left( {1 + \sin x} \right)^{\frac{1}{2}}} + c\\I = 2\sqrt {1 + \sin x} + c\end{array}\]

Question (29)

\[\cot x\log \left( {\sin x} \right)\]

solution:

Let log(sinx) = t
Taking derivative
\[\begin{array}{l}\frac{1}{{\sin x}} \cdot \cos xdx = dt\\\cot xdx = dt\end{array}\] Substituting values in equation \[\begin{array}{l}I = \int {tdt} \\I = \frac{{{t^2}}}{2} + c\\I = \frac{1}{2}{\left[ {\log \left( {\sin x} \right)} \right]^2} + c\end{array}\]

Question (30)

\[\frac{{\sin x}}{{{{\left( {1 +cos x} \right)}}}}\]

solution:

\[I = \int {\frac{{\sin x}}{{\left( {1 + cosx} \right)}}dx} \] Let 1+cosx = t
Taking derivative
∴ -sinx dx = dt
sinxdx = -dt
substituting values in given equation \[\begin{array}{l}I = \int {\frac{{ - dt}}{t}} \\I = - \log \left| t \right| + c\\I = - \log \left| {1 + \cos x} \right| + c\end{array}\]

Question (31)

\[\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}\]

solution:

\[I = \int {\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx} \] Let 1+cosx = t
∴ -sinxdx = dt
sinxdx = -dt
On substituting above values in given equation \[\begin{array}{l}I = \int { - \frac{{dt}}{{{t^2}}}} \\I = - \int {{t^{ - 2}}dt} \\I = - \frac{{{t^{ - 1}}}}{{ - 1}} + c\\I = \frac{1}{t} + c\\I = \frac{1}{{1 + \cos x}} + c\end{array}\]

Question (32)

\[\frac{1}{{1 + \cot x}}\]

solution:

\[\begin{array}{l}I = \int {\frac{1}{{1 + \cot x}}dx} \\I = \int {\frac{1}{{1 + \frac{{\cos x}}{{\sin x}}}}} dx\\I = \int {\frac{{\sin x}}{{\sin x + \cos x}}} dx\end{array}\]
We will try write numerator such way that one part will be derivative of denominator and one part is denominator so that we can separate it
\[\begin{array}{l}I = \frac{1}{2}\int {\frac{{2\sin x}}{{\sin x + \cos x}}} \\I = \frac{1}{2}\int {\frac{{\sin x + \cos x + sinx - \cos x}}{{\sin x + \cos x}}} dx\\I = \frac{1}{2}\left[ {\int {\frac{{\sin x + \cos x}}{{\sin x + \cos x}} + \frac{{\sin x - \cos x}}{{\sin x + \cos x}}} } \right]dx\\I = \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}dx} \end{array}\]

Let sinx + cost = t
∴ (cosx-sinx) dx = dt
(sinx -cosx) = -dt
Substituting values
\[\begin{array}{l}I = \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{{ - dt}}{t}} \\I = \frac{x}{2} - \frac{1}{2}\log \left| t \right| + c\\I = \frac{x}{2} - \frac{1}{2}\log \left| {\sin x + \cos x} \right| + c\end{array}\]

Question (33)

\[\frac{1}{{1 - tanx}}\]

solution:

\[\begin{array}{l}I = \int {\frac{1}{{1 - tanx}}dx} \\I = \int {\frac{1}{{1 - \frac{{\sin x}}{{\cos x}}}}dx} \\I = \int {\frac{{\cos x}}{{\cos x - \sin x}}} dx\\I = \frac{1}{2}\int {\frac{{2\cos x}}{{\cos x - \sin x}}} dx\\I = \frac{1}{2}\int {\frac{{\cos x - \sin x + \cos x + \sin x}}{{\cos x - \sin x}}} \\I = \frac{1}{2}\left[ {\int {\frac{{\cos x - \sin x}}{{\cos x - \sin x}}dx + \int {\frac{{\cos x + \sin x}}{{\cos x - \sin x}}dx} } } \right]\\I = \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{{\cos x + \sin x}}{{\cos x - \sin x}}dx} \end{array}\] cosx - sinx =t
(-sinx-cosx) dx = dt
(sinx+cosx) dx = dt
On substitution \[\begin{array}{l}I = \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{{\cos x + \sin x}}{{\cos x - \sin x}}dx} \\I = \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{{ - dt}}{t}} \\I = \frac{x}{2} - \frac{1}{2}\log \left| t \right| + c\\I = \frac{x}{2} - \frac{1}{2}\log \left| {\cos x - \sin x} \right| + c\end{array}\]

Question (34)

\[\frac{{\sqrt {\tan x} }}{{\sin x\cos x}}\]

solution:

\[I = \int {\frac{{\sqrt {\tan x} }}{{\sin x\cos x}}dx} \]
Here we can not assume sinxcox as t as its derivative is not as numerator
If we assume tanx = t2 then its derivative is sec2, which also not there is numerator , get sec2 in numerator divide numerator and denominator by cos2x
\[\begin{array}{l}I = \int {\frac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx\\I = \int {\frac{{\frac{{\sqrt {\tan x} }}{{{{\cos }^2}x}}}}{{\frac{{\sin x\cos x}}{{{{\cos }^2}x}}}}} dx\\I = \int {\frac{{\sqrt {\tan x} {{\sec }^2}x}}{{\tan x}}} dx\\I = \int {\frac{{{{\sec }^2}x}}{{\sqrt {\tan x} }}} dx\end{array}\] let tanx = t2
⇒ t= √tanx
∴ sec2x dx = 2t dt
\[\begin{array}{l}I = \int {\frac{{2t}}{{\sqrt {{t^2}} }}dt} \\I = 2\int {dt} \\I = 2t + c\\I = 2\sqrt {\tan x} + c\end{array}\]

Question (35)

\[\frac{{{{\left( {1 + \log x} \right)}^2}}}{x}\]

solution:

\[I = \int {\frac{{{{\left( {1 + \log x} \right)}^2}}}{x}dx} \] Let (1+logx) = t
$$\therefore \frac{1}{x}dx=dt$$ $$\begin{array}{l}I = \int {{{\left( {1 + \log x} \right)}^2}\frac{1}{x}} dx\\I = {\int t ^2}dt\\I = \frac{{{t^3}}}{3} + c\\I = \frac{{{{\left( {1 + \log x} \right)}^3}}}{3} + c\end{array}$$

Question (36)

\[\frac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}\]

solution:

\[\begin{array}{l}I = \int {\frac{{\left( {x + 1} \right){{\left( {x + \log x} \right)}^2}}}{x}} \\I = \int {\left( {\frac{{x + 1}}{x}} \right)} {\left( {x + \log x} \right)^2}\\I = \int {\left( {1 + \frac{1}{x}} \right)} {\left( {x + \log x} \right)^2}\end{array}\] Let (x+logx)=t
$$\therefore \left( {1 + \frac{1}{x}} \right)dx = dt$$ \[\begin{array}{l}I = \int {{{\left( {x + \log x} \right)}^2}\left( {1 + \frac{1}{x}} \right)dx} \\I = \int {{t^2}} dt\\I = \frac{{{t^3}}}{3} + c\\I = \frac{{{{\left( {x + \log } \right)}^3}}}{3} + c\end{array}\]

Question (37)

\[{\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}}\]

solution:

\[I = \int {\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}dx} \]
Normally composite function is taken as t
Here sin(tan-1x4) is composite but its derivative as (tan-1x4). Its term is not numerator
Let tan-1x4=t
Diff. w.r.t. x, we get,
$$\therefore\frac{1}{{1 + {{\left( {{x^4}} \right)}^2}}} \times \frac{d}{{dx}}{x^4} = \frac{{dt}}{{dx}}$$ \[\begin{array}{l}\frac{1}{{1 + {x^8}}} \cdot 4{x^3}dx = dt\\\frac{{{x^3}}}{{1 + {x^8}}}dx = \frac{{dt}}{4}\end{array}\] \[{I = \int {\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}} dx}\] \[{I = \int {\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)\frac{{{x^3}}}{{1 + {x^8}}}} dx}\] \[{I = \frac{1}{4}\int {\sin tdt} }\] \[{I = - \frac{1}{4}\cos t + c}\] \[{I = - \frac{1}{4}\cos \left( {{{\tan }^{ - 1}}{x^4}} \right) + c}\]

Question (38)

\[\int {\frac{{10{x^9} + {{10}^x}\log {e^{10}}}}{{{x^{10}} + {{10}^x}}}dx} \] Options
(A)10x - x10 +c
(B) 10x + x10 +c
(C)(10x - x10)-1 +c
(D) log( 10x + x10) + c

solution:

\[I = \int {\frac{{10{x^9} + {{10}^x}\log {e^{10}}}}{{{x^{10}} + {{10}^x}}}dx} \] Let x10+10x = t
differentiating w.r.t "t"
Formula \[\frac{d}{{dx}}{a^x} = {a^x}\log _e^a\]
(10x9 +10xloge10)dx=dt
\[\begin{array}{l}I = \int {\frac{{dt}}{t}} \\I = \log \left| t \right| + c\\I = \log \left| {{x^{10}} + {{10}^x}} \right| + c\end{array}\] So D is correct option

Question (39)

$$\int {\frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}} \text(equals)$$ (A) tanx + cotx + c
(B) tanx - cotx + c
(C) tanxcotx + c
(D) tanx - cot2x + c

solution:

\[I = \int {\frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}} \] Replace numerator using identity sin2θ + cos2θ =1
\[\begin{array}{l}I = \int {\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} \\I = \int {\left[ {\frac{{{{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} \right]} dx\\I = \int {\frac{1}{{{{\cos }^2}x}}dx + \int {\frac{1}{{{{\sin }^2}x}}dx} } \\I = \int {{{\sec }^2}xdx + \int {\cos e{c^2}xdx} } \\I = \tan x - \cot x + c\end{array}\] Option (B) is correct
Exercise7.1 ⇐
⇒ Exercise7.3