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: \[I = \int {\frac{{2x}}{{1 + {x^2}}}dx} \] Let 1+x² = t
Differentiate w.r.t. x 2xdx = dt
Replacing value of 2xdx and 1+x² we get \[\begin{array}{l}I = \int {\frac{{2x}}{{1 + {x^2}}}dx} \\I = \int {\frac{{dt}}{t}} \end{array}\] 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: \[I = \int {\sin x\sin \left( {\cos x} \right)} dx\] 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:\[I = \int {\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx} \] \[\begin{array}{l}I = \frac{1}{2}\int {2\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx} \\I = \frac{1}{2}\int {\sin 2\left( {ax + b} \right)dx} \end{array}\] \[\begin{array}{l}I = \frac{1}{2}\frac{{ - \cos 2\left( {ax + b} \right)}}{{2a}} + c\\I = \frac{{ - \cos 2\left( {ax + b} \right)}}{{4a}} + c\end{array}\] 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"
\[\begin{array}{l}I = \frac{{{t^2}}}{{2a}} + c'\\I = \frac{{{{\sin }^2}\left( {ax + b} \right)}}{{2a}} + c'\end{array}\] \[\begin{array}{l}I = \frac{{{t^2}}}{{2a}} + c'\\I = \frac{1}{{2a}}\left[ {\frac{{1 - \cos 2\left( {ax + b} \right)}}{2}} \right] + c'\\I = \frac{1}{{4a}} - \frac{1}{{4a}}\cos 2\left( {ax + b} \right) + c'\\let\left[ {c = c' + \frac{1}{{4a}}} \right]\\I = - \frac{1}{{4a}}\cos 2\left( {ax + b} \right) + c\end{array}\]

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 \[\begin{array}{l}I = \int {\frac{{\frac{{dt}}{2}}}{{{e^t}}}} \\I = \frac{1}{2}\int {{e^{ - t}}} dt\\I = \frac{1}{2} \cdot \frac{{{e^{ - t}}}}{{ - 1}} + c\\I = - \frac{1}{{2{e^t}}} + c\\I = - \frac{1}{{2{e^{{x^2}}}}} + c\end{array}\]

Question (18)

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

solution

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

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 e^x \[I = \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} \] Let e^x+e^-x = t
∴ [e^x+e^-x(-1)]dx = dt
∴ (e^x-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 e^2x+e^-2x=t
[e^2x.(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

\[I = \int {{{\tan }^{ 2}}\left( {2x - 3} \right)dx} \] \[\begin{array}{l}I = \int {\left[ {{{\sec }^2}\left( {2x - 3} \right) - 1} \right]} dx\\I = \int {{{\sec }^2}\left( {2x - 3} \right)dx - \int {dx} } \end{array}\] 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 \[\begin{array}{l}I = \int {\frac{{ - dt}}{t}} \\I = - \log \left| t \right| + c\\I = - \log \left| {1 + \cos x} \right| + c\end{array}\]