12th NCERT INTEGRALS Exercise 7.3 Questions 24
Do or do not
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Integrate the functions in Exercise 1 to 22

Question (1)

\[{\sin ^2}\left( {2x + 5} \right)\]

Solution

There is no formula for integration of sin2, So first we use formula \[{\sin ^2}x = \frac{{1 - \cos 2x}}{2}\] to separate terms for integration
$$\begin{array}{l}I = \int {{{\sin }^2}\left( {2x + 5} \right)dx} \\I = \int {\frac{{1 - \cos 2\left( {2x + 5} \right)}}{2}dx} \end{array}$$ \[I = \int {\frac{1}{2}dx - \int {\frac{{\cos \left( {4x + 10} \right)}}{2}dx} } \]
as 4x-10 is linear function
\[\int {\cos \left( {4x + 10} \right)dx = \frac{{\sin \left( {4x + 10} \right)}}{{\frac{d}{{dx}}\left( {4x + 10} \right)}}} \]
\[\begin{array}{l}I = \frac{x}{2} - \frac{1}{2}\frac{{\sin \left( {4x + 10} \right)}}{4} + c\\I = \frac{x}{2} - \frac{{\sin \left( {4x + 10} \right)}}{8} + c\end{array}\]

Question (2)

\[\sin 3x\cos 4x\]

Solution

As there is no formula for multiplication we will write it in the form of addition or subtraction form . We can use trigonometric formula
2sinAcosB = sin(A+B) + sin(A-B) As there is no formula for multiplication we will write it in the form of addition or subtraction form . We can use trigonometric formula
2sinAcosB = sin(A+B) + sin(A-B)
\[\begin{array}{l}I = \int {\sin 3x\cos 4xdx} \\I = \frac{1}{2}\int 2 \sin 3x\cos 4dx\end{array}\] \[I = \frac{1}{2}\int {\left[ {\sin \left( {3x + 4x} \right) + \sin \left( {3x - 4x} \right)} \right]} dx\] \[\begin{array}{l}I = \frac{1}{2}\int {\left[ {\sin 7x + \sin \left( { - x} \right)} \right]} dx\\I = \frac{1}{2}\int {\left( {\sin 7x - \sin x} \right)} dx\\I = \frac{1}{2}\int {\sin 7xdx} - \frac{1}{2}\int {\sin xdx} \end{array}\]
sin(-θ) = -sinθ
\[\int {\sin \left( {ax} \right)} dx = \frac{{ - \cos \left( {ax} \right)}}{{\frac{d}{{dx}}\left( {ax} \right)}}\]
\[I = \frac{1}{2}\frac{{\left( { - \cos 7x} \right)}}{7} - \frac{1}{2}\left( { - \cos x} \right) + C\] \[I = \frac{{\cos x}}{2} - \frac{{\cos 7x}}{{14}} + c\]

Question (3)

\[\cos 2x\cos 4x\cos 6x\]

Solution

\[I = \int {\cos 2x\cos 4x\cos 6xdx} \]
Use following formula to simplify the given equation
2cosAcosB = cos(A+B)+cos(A-B) multiply and divide equation by 2
\[I = \frac{1}{2}\int {2 \cdot \cos 2x\cos 4x\cos 6xdx} \] \[I = \frac{1}{2}\int {\cos 4x\left( {2\cos 2x\cos 6x} \right)dx} \] \[I = \frac{1}{2}\int {\cos 4x\left[ {\cos \left( {2x + 6x} \right) + \cos \left( {2x - 6x} \right)} \right]} dx\] \[I = \frac{1}{2}\int {\cos 4x\left[ {\cos 8x + \cos \left( { - 4x} \right)} \right]} dx\]
\[{\cos ^2}A = \frac{{1 + \cos 2A}}{2}\] 2cosAcosB = cos(A+B)+ cos (A-B)
Note cos(-θ) = -cosθ \[I = \frac{1}{2}\int {\cos 4x} \left[ {\cos 8x + \cos 4x} \right]dx\] \[I = \frac{1}{2}\left( {\int {\cos 4x\cos 8x} + {{\cos }^2}4x} \right)dx\]
2cosAcosB= cos(A+b)+cos(A-B) \[{\cos ^2}A = \frac{{1 + \cos 2A}}{2}\]
\[I = \frac{1}{2}\int {\left[ {\frac{{\cos \left( {4x + 8x} \right) + \cos (4x - 8x)}}{2} + \frac{{1 + \cos 8x}}{2}} \right]} dx\] \[I = \frac{1}{4}\int {\cos 12xdx + \frac{1}{4}\int {\cos \left( { - 4x} \right)} dx + \int {\frac{1}{4}dx} + \frac{1}{4}\int {\cos 8xdx} } \] Note cos(-A) = +cosA \[I = \frac{1}{4}\frac{{\sin 12x}}{{12}} + \frac{1}{4}\frac{{\sin 4x}}{4} + \frac{x}{4} + \frac{1}{4}\frac{{\sin 8x}}{8} + c\]

Question (4)

\[{\sin ^3}\left( {2x + 1} \right)\]

Solution

Method I
First split sin3 as sinxsin2x and replace by sin2x = 1-cos2x
\[I = \int {\sin \left( {2x + 1} \right){{\sin }^2}\left( {2x + 1} \right)dx} \] \[I = \int {\left[ {1 - {{\cos }^2}(2x+1)} \right]\sin \left( {2x + 1} \right)dx} \] Let cos(2x+1) = t
-sin(2x+1).2dx= dt \[\sin \left( {2x + 1} \right)dx = \frac{{dt}}{{ - 2}}\] substituting values \[\begin{array}{l}I = \int {\left( {1 - {t^2}} \right)\left( {\frac{{ - dt}}{2}} \right)} \\I = - \frac{1}{2}\int t + \frac{1}{2}\int {{t^2}dt} \\I = - \frac{t}{2} + \frac{1}{2}\frac{{{t^3}}}{3} + c\\I = - \frac{t}{2} + \frac{{{t^3}}}{6} + c\end{array}\] \[I = - \frac{{\cos \left( {2x + 1} \right)}}{2} + \frac{{{{\cos }^3}\left( {2x + 1} \right)}}{6} + c\] Method II
we will use sin3x=3sinx - 4sin3x \[{\sin ^3}x = \frac{{ - \sin 3x + 3\sin x}}{4}\]
\[I = \int {{{\sin }^3}\left( {2x + 1} \right)dx} \] \[I = \int {\frac{{ - \sin 3\left( {2x + 1} \right) + 3\sin \left( {2x + 1} \right)}}{4}dx} \] \[I = \frac{1}{4}\int { - \sin \left( {6x + 3} \right)dx + \frac{3}{4}\int {\sin \left( {2x + 1} \right)dx} } \] \[I = \frac{{ - 1}}{4}\left( {\frac{{ - \cos \left( {6x + 3} \right)}}{6}} \right) + \frac{3}{4}\left( {\frac{{ - \cos \left( {2x + 1} \right)}}{2}} \right) + c\] \[I = \frac{{\cos 3(2x + 1)}}{{24}} - \frac{{3\cos (2x + 1)}}{8} + c\] Answers of Method I and Method II looks different but are same
use formula for cos3x = 4cos3x -3cosx. for second method answer
\[I = \frac{{4{{\cos }^3}(2x + 1) - 3\cos (2x + 1)}}{{24}} - \frac{{3\cos (2x + 1)}}{8} + c\] \[I = \frac{{{{\cos }^3}(2x + 1)}}{6} - \frac{{\cos (2x + 1)}}{8} - \frac{{3\cos (2x + 1)}}{8} + c\] Is the answer given in Method I

Question (5)

\[{\sin ^3}x{\cos ^3}xdx\]

Solution

In such case odd power is break into multiplication of two terms. If both are odd any one is split up
\[I = \int {{{\sin }^3}x{{\cos }^3}xdx} \] \[I = \int {{{\sin }^3}x{{\cos }^2}x\cos xdx} \] \[I = \int {{{\sin }^3}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} \] Let sinx = t
cosxdx = dt
Substitute above values in given equation \[I = \int {{t^3}\left( {1 - {t^2}} \right)dt} \] \[\begin{array}{l}I = \int {{t^3}dt - } \int {{t^5}dt} \\I = \frac{{{t^4}}}{4} - \frac{{{t^6}}}{6} + c\\I = \frac{{{{\sin }^4}x}}{4} - \frac{{{{\sin }^6x}}}{6} + c\end{array}\] Method II \[\begin{array}{l}I = \int {{{\sin }^3}x{{\cos }^3}xdx} \\I = \int {{{\sin }^2}x\sin x{{\cos }^3}xdx} \\I = \int {(1 - {{\cos }^2}x){{\cos }^3}x\sin xdx} \end{array}\] Let cosx = t
-sinx dx = dt
sinx dx = - dt
Substituting in given equation \[\begin{array}{l}I = \int {(1 - {t^2}){t^3}\left( { - dt} \right)} \\I = - \int {\left( {{t^3} - {t^5}} \right)} dt\\I = - \left( {\frac{{{t^4}}}{4} - \frac{{{t^6}}}{6}} \right) + c\\I = - \frac{{{t^4}}}{4} + \frac{{{t^6}}}{6} + c\end{array}\] Substitute value of "t" \[I = - \frac{{{{\cos }^4}x}}{4} + \frac{{{{\cos }^6}}}{6} + c\]

Question (6)

sinx sin2x sin3x

Solution

\[I = \int {\sin x\sin 2x\sin 3xdx} \]
I given question, it is a multiplication of same function. So first separate in addition or subtraction. Use formula
-2sinA sinB = cos (A+B) - cos(A-B)
2sinθcosθ = sin2θ
2cosA sinB = sin(A+B) - sin(A-B)
sin(-θ) = - sinθ
cos(-θ) = cosθ
\[I = - \frac{1}{2}\int {\left( { - 2\sin x\sin 2x} \right)\sin 3xdx} \] \[I = - \frac{1}{2}\int {\left[ {\cos (x + 2x) - \cos (x - 2x)} \right]} \cdot \sin 3xdx\] \[I = - \frac{1}{2}\int {\left[ {\cos 3x - cos( - x)} \right]} \cdot \sin 3xdx\] \[I = - \frac{1}{2}\int {\left[ {\cos 3x - \cos x)} \right]} \cdot \sin 3xdx\] \[I = -\frac{1}{2}\int {\cos 3x\sin 3xdx + \frac{1}{2}\int {\cos x\sin 3xdx} } \] \[I = - \frac{1}{4}\int {2\cos 3x\sin 3xdx + \frac{1}{4}\int {2\cos x\sin 3xdx} } \] \[I = - \frac{1}{4}\int {\sin 6xdx + \frac{1}{4}\int {\left[ {\sin (x + 3x) - \sin (x - 3x)} \right]dx} } \] \[I = - \frac{1}{4}\int {\sin 6xdx + \frac{1}{4}\int {\sin 4xdx + \frac{1}{4}\int {\sin 2xdx} } } \] \[I = \frac{{\cos 6x}}{{24}} - \frac{{\cos 4x}}{{16}} - \frac{{\cos 2x}}{8} + c\]

Question (7)

sin4x sin8x

Solution

\[I = \int {\sin 4x\sin 8xdx} \]
It is multiplication of same function. So first separate in addition or subtraction. For it we will use formula
-2sin A cosB = cos(A+B) - cos(A-B)
cos(-x) = cosx
\[\int {\cos \left( {ax} \right)dx = } \frac{{\sin ax}}{a} + c\]
\[I = - \frac{1}{2}\int { - 2\sin 4x\sin 8xdx} \] \[= - \frac{1}{2}\int {\left[ {\cos \left( {4x + 8x} \right) - \cos \left( {4x - 8x} \right)} \right]} dx\] \[ = - \frac{1}{2}\int {\cos 12xdx + \frac{1}{2}\int {\cos 4xdx} } \] \[ = - \frac{1}{2}\frac{{\sin 12x}}{{12}} + \frac{1}{2}\frac{{\sin 4x}}{4} + c\] \[ = \frac{{\sin 4x}}{8} - \frac{{\sin 12x}}{{24}} + c\]

Question (8)

\[\frac{{1 - \cos x}}{{1 + \cos x}}\]

Solution

\[I = \int {\frac{{1 - \cos x}}{{1 + \cos x}}dx} \]
There is no formula for division so we will simplify it first. Formula used
1- cos2x = 2sin2x
1 + cos2x = 2cos2
tan2 = sec2 - 1
\[I = \int {\frac{{2{{\sin }^2}\frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} dx\] \[I = \int {{{\tan }^2}\frac{x}{2}} dx\] as tan2θ = sec2θ - 1
\[I = \int {\left( {{{\sec }^2}\frac{x}{2} - 1} \right)dx} \]
\[\int {{{\sec }^2}\left( {\frac{x}{a}} \right)dx = \frac{{\tan \left( {\frac{x}{a}} \right)}}{{\frac{1}{a}}}} \]
\[I = \frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}} - x + c\] \[I = 2\tan \frac{x}{2} - x + c\]

Question (9)

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

Solution

\[I = \int {\frac{{\cos x}}{{1 + \cos x}}dx} \] No formula for division so given equation must be simplified \[I = \int {\frac{{1 + \cos x - 1}}{{1 + \cos x}}dx} \] \[ = \int {\frac{{1 + \cos x}}{{1 + \cos x}}dx - \int {\frac{1}{{1 + \cos x}}dx} } \]
1+cosθ = 2cos2(θ/2)
\[ = \int {dx} - \int {\frac{1}{{2{{\cos }^2}\frac{x}{2}}}dx} \] \[ = \int {dx} - \frac{1}{2}\int {{{\sec }^2}\frac{x}{2}} \] \[ = x - \frac{1}{2}\frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}} + c\] \[ = x - \tan \frac{x}{2} + c\] OR Method II: We can rationalize the denominator and them separate
\[I = \int {\frac{{\cos x}}{{1 + \cos x}}} dx\] \[ = \int {\frac{{\cos x(1 - \cos x)}}{{(1 + \cos x)(1 - \cos x)}}} \] \[ = \int {\frac{{\cos x - {{\cos }^2}x}}{{1 - {{\cos }^2}x}}} \] \[{ = \int {\frac{{\cos x - {{\cos }^2}x}}{{{{\sin }^2}x}}} }\] \[ = \int {\frac{{\cos x}}{{{{\sin }^2}x}} - \int {{{\cot }^2}x} } \] \[ = \int {\cot x\cos ecxdx - \int {{{\cot }^2}x} } \] Note: cot2θ = cosec2θ - 1 \[ = \int {\cot x\cos ecxdx - \int {(\cos e{c^2}x - 1)dx} } \] \[ = \int {\cot x\cos ecxdx - \int {\cos e{c^2}xdx - \int {dx} } } \] \[\begin{array}{l} = - \cos ecx - ( - \cot x) + x + c\\ = - \cos ecx + \cot x + x + c\end{array}\]

Question (10)

sin4x dx

Solution

\[I = \int {{{\sin }^4}xdx} \]
Power of sin is even so we can not break into sin . sin
We can use following formula to simplify \[{\sin ^2}x = \frac{{1 - \cos 2x}}{2}\] and (a - b)2 = a2 -2ab + b2
\[I = \int {{{\left( {{{\sin }^2}x} \right)}^2}dx} \] \[I = \int {{{\left( {\frac{{1 - \cos 2x}}{2}} \right)}^2}dx} \] \[I = \int {\frac{{1 - 2\cos 2x + {{\cos }^2}2x}}{4}dx} \] \[ = \frac{1}{4}\int {dx - \frac{2}{4}\int {\cos 2xdx + \frac{1}{4}\int {{{\cos }^2}2xdx} } } \]
use identity \[{\cos ^2}a = \frac{{1 + \cos 2a}}{2}\]
\[ = \frac{1}{4}\int {dx - \frac{2}{4}\int {\cos 2xdx + \frac{1}{4}\int {\frac{{1 + \cos 4x}}{2}dx} } } \] \[ = \frac{1}{4}\int {dx - \frac{2}{4}\int {\cos 2xdx + \frac{1}{4}\int {\frac{1}{2}dx} } } + \frac{1}{8}\int {\frac{{\cos 4x}}{1}} dx\] \[ = \frac{x}{4} - \frac{{\sin 2x}}{4} + \frac{x}{8} + \frac{{\sin 4x}}{{32}} + c\] \[ = \frac{{3x}}{8} - \frac{{\sin 2x}}{4} + \frac{{\sin 4x}}{{32}} + c\]

Question (11)

cos42x

Solution

\[I = \int {{{\cos }^4}2xdx} \] \[I = {\int {\left( {{{\cos }^2}2x} \right)} ^2}dx\] \[ = \int {{{\left( {\frac{{1 + \cos 4x}}{2}} \right)}^2}dx} \] \[ = \int {\frac{{1 + 2\cos 4x + {{\cos }^2}4x}}{4}dx} \] \[ = \frac{1}{4}\int {dx} + \frac{1}{4}\int {2\cos 4x dx + \frac{1}{4}} \int {{{\cos }^2}4xdx} \] \[ = \frac{1}{4}\int {dx} + \frac{1}{4}\int {2\cos 4x dx+ \frac{1}{4}} \int {\frac{{1 + \cos 8x}}{2}} \] \[{ = \frac{1}{4}\int {dx} + \frac{1}{4}\int {2\cos 4xdx + \frac{1}{4}\int {\frac{{dx}}{2} + \frac{1}{4}\int {\frac{{\cos 8x}}{2}dx} } } }\] \[ = \frac{x}{4} + \frac{{\sin 4x}}{8} + \frac{x}{8} + \frac{1}{8}\frac{{\sin 8x}}{8} + c\] \[ = \frac{x}{4} + \frac{{\sin 4x}}{8} + \frac{x}{8} + \frac{{\sin 8x}}{{64}} + c\]

Question (12)

\[\frac{{{{\sin }^2}x}}{{1 + \cos x}}\]

Solution

\[I = \int {\frac{{{{\sin }^2}x}}{{1 + \cos x}}dx} \] Replace sin2x = 1 - cos2x \[I = \int {\frac{{1 - {{\cos }^2}x}}{{1 + \cos x}}dx} \] \[I = \int {\frac{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}{{1 + \cos x}}} dx\] \[I = \int {\left( {1 - \cos x} \right)dx} \] \[I = \int {dx} - \int {\cos xdx} \] \[I = x - \sin x + c\]

Question (13)

\[\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}\]

Solution

\[I = \int {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}dx} \]
we will use cos2a= 2cos2a - 1 for numerator to get cos terms as cosx and cosα
\[I = \int {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}dx} \] \[ = \int {\frac{{\left( {2{{\cos }^2}x - 1} \right) - \left( {2{{\cos }^2}\alpha - 1} \right)}}{{\cos x - \cos \alpha }}dx} \] \[ = 2\int {\frac{{{{\cos }^2}x - {{\cos }^2}\alpha }}{{\cos x - \cos \alpha }}} dx\] \[ = 2\int {\left( {\cos x + \cos \alpha } \right)} dx\] Note α is constant \[ = 2\sin x + 2\cos \alpha \cdot x + c\]

Question (14)

\[\frac{{\cos x - \sin x}}{{1 + \sin 2x}}\]

Solution

Here angle in numerator is x while in denominator it is 2x, try bring both in same that is x,. We can use following
1 + sin2x =1+2sinxcosx
1 + sin2x = sin2x + cos2 + 2sinxcosx
1 + sin2x = (sinx+cosx)2
\[I = \int {\frac{{\cos x - \sin x}}{{1 + \sin 2x}}dx} \] \[ = \int {\frac{{\cos x - \sin x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}dx} \] Let sinx+cosx = t
∴ (cosx -sinx) dx = dt
\[I = \int {\frac{{dt}}{{{t^2}}}} \] \[I = \int {{t^{ - 2}}dt} \] \[I = \frac{{{t^{ - 1}}}}{{ - 1}} + c\] \[I = - \frac{1}{t} + c\] \[I = - \frac{1}{{\sin x + \cos x}} + c\]

Question (15)

tan32x sec2x

Solution

\[I = \int {{{\tan }^3}2x\sec 2xdx} \] \[I = \int {{{\tan }^2}2x} \tan 2x\sec 2xdx\]
1+tan2a = sec2
\[I = \int {\left( {{{\sec }^2}2x - 1} \right)} \tan 2x\sec 2xdx\] Let sec2x = t
sec2x tan2x 2dx = dt
sec2x tan2x dx = dt/2
Substituting above value in given equation
\[I = \int {\left( {{t^2} - 1} \right)\frac{{dt}}{2}} \] \[I = \frac{1}{2}\int {{t^2}dt} - \frac{1}{2}\int {dt} \] \[I = \frac{1}{2}\frac{{{t^3}}}{3} - \frac{1}{2}t + c\] \[I = \frac{1}{6}{\sec ^3}2x - \frac{{\sec 2x}}{2} + c\]

Question (16)

tan4

Solution

\[I = \int {{{\tan }^4}xdx} \] \[I = \int {{{\tan }^2}x{{\tan }^2}xdx} \]
One of tan2 will be replace by sec2x - 1.
\[I = \int {{{\tan }^2}x\left( {{{\sec }^2}x - 1} \right)dx} \] \[I = \int {{{\tan }^2}x{{\sec }^2}x - \int {{{\tan }^2}xdx} } \] \[I = \int {{{\tan }^2}x{{\sec }^2}x - \int {\left( {{{\sec }^2}x - 1} \right)dx} } \] Let tan x = t
sec2x = dt \[I = \int {{t^2}dt - \int {{{\sec }^2}xdx + \int {dx} } } \] \[I = \frac{{{t^3}}}{3} - \tan x + x + c\] \[I = \frac{1}{3}{\tan ^3}x - \tan x + x + c\]

Question (17)

\[\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}\]

Solution

\[I = \int {\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} \] \[ = \int {\frac{{{{\sin }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx + \int {\frac{{{{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} \] \[ = \int {\tan x\sec xdx + \int {\cot x\cos ecxdx} } \] \[ = \sec x - \cos ecx + c\]

Question (18)

\[\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}\]

Solution

\[I = \int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} \]
Use formula 2sin2x = 1- cos2x to replace sin2 term
\[I = \int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} \] \[I = \int {\frac{{1 - 2{{\sin }^2}x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} dx\] \[I = \int {\frac{1}{{{{\cos }^2}x}}} dx\] \[I = \int {{{\sec }^2}xdx} \] \[I = \tan x + c\]

Question (19)

\[\frac{1}{{\sin x{{\cos }^3}x}}\]

Solution

\[I = \int {\frac{1}{{\sin x{{\cos }^3}x}}dx} \]
We will divide numerator and denominator by cos4. to get sec4x in denominator then use sec2x = 1+tan2x
Then by taking derivative of tanx to get secx
\[I = \int {\frac{{\frac{1}{{{{\cos }^4}x}}}}{{\frac{{\sin x{{\cos }^3}x}}{{{{\cos }^4}x}}}}} dx\] \[I = \int {\frac{{{{\sec }^4}x}}{{\tan x}}dx} \] \[I = \int {\frac{{{{\sec }^2}x{{\sec }^2}x}}{{\tan x}}dx} \] \[I = \int {\frac{{\left( {1 + {{\tan }^2}x} \right){{\sec }^2}x}}{{\tan x}}dx} \] Let tan x = t
sec2xdx= dt
Substituting above value in equation \[I = \int {\frac{{\left( {1 + {t^2}} \right)}}{t}dt} \] \[I = \int {\frac{1}{t}dt + \int {tdt} } \] \[I = \log \left| t \right| + \frac{{{t^2}}}{2} + c\] Substituting value of t \[I = \log \left| {\tan x} \right| + \frac{1}{2}{\tan ^2}x + c\]

Question (20)

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

Solution

\[I = \int {\frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx} \] Replace cos2x by the formula cos2x-sin2x
\[I = \int {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}} dx\] \[I = \int {\frac{{\left( {\cos x + \sin x} \right)\left( {\cos x - sinx} \right)}}{{{{\left( {\cos x + \sin x} \right)}^2}}}} dx\] \[I = \int {\frac{{\cos x - sinx}}{{\cos x + sinx}}} dx\] Let cosx + sinx = t
∴ (-sinx+cosx)dx = dt
On substitution \[I = \int {\frac{{dt}}{t}} \] \[I = \log \left| t \right| + c\] \[I = \log \left| {\cos x + \sin x} \right| + c\]

Question (21)

sin-1cosx

Solution

\[I = \int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx} \]
We know that sin-1 (sinθ) = θ
We will replace cosx in terms of sinx using following formula \[\cos x = \sin \left( {\frac{\pi }{2} - x} \right)\]
\[I = \int {{{\sin }^{ - 1}}\left[ {\sin \left( {\frac{\pi }{2} - x} \right)} \right]} dx\] \[I = \int {\left( {\frac{\pi }{2} - x} \right)} dx\] \[I = \frac{\pi }{2}x - \frac{{{x^2}}}{2} + c\]

Question (22)

\[\frac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}\]

Solution

\[I = \int {\frac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}dx} \]
As denominator have both terms are cos, we will represent them in sin
we will divide and multiply equation by sin (b-a) which can be represented as sin [(x-a) -(x-b)] and use formula sin (A-B) = sinA cosB - cosA sinB
\[I = \frac{1}{{\sin (b - a)}}\int {\frac{{\sin \left[ {(x - a) - (x - b)} \right]}}{{\cos (x - a)\cos (x - b)}}dx} \] \[ = \frac{1}{{\sin \left( {b - a} \right)}}\int {\frac{{\sin \left( {x - a} \right)}}{{\cos \left( {x - a} \right)}}} dx - \frac{1}{{\sin \left( {b - a} \right)}}\int {\frac{{\sin \left( {x - b} \right)}}{{\cos \left( {x - b} \right)}}} dx\]
\[\int {\tan xdx = \int {\frac{{\sin x}}{{\cos x}}dx = - \log \left| {\cos x} \right|} } + c\]
\[ = \frac{1}{{\sin \left( {b - a} \right)}}\left[ { - \log \left| {\cos \left( {x - a} \right)} \right| - \left( { - \log \left| {\cos \left( {x - b} \right)} \right|} \right)} \right] + c\] \[ = \frac{1}{{\sin \left( {b - a} \right)}}\log \left| {\frac{{\cos \left( {x - a} \right)}}{{\cos \left( {x - b} \right)}}} \right| + c\]

Question (23)

\[\int {\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} \] is equal to (A) tan x + cot x + c
(B) tan x + cosec x + c
(C) -tan x + cot x + c
(D) tan x + sec x + c

Solution

\[I = \int {\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} \] \[I = \int {\frac{{{{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx - \int {\frac{{{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} \] \[ = \int {{{\sec }^2}xdx - \int {\cos e{c^2}xdx} } \] \[ = \tan x - \left( { - \cot x} \right) + c\] = tanx + cotx + c
Option (A) is correct

Question (23)

\[\int {\frac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {x{e^x}} \right)}}} \] is equal to (A) -cot (exx) + c
(B) tan (xxx) + c (C) tan (ex) + c (D) cot (ex) + c

Solution

\[I = \int {\frac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {x{e^x}} \right)}}} \] Let xex = t
Differentiating w.r.t x
\[x{e^x} + {e^x}\left( 1 \right) = \frac{{dt}}{{dx}}\] \[{e^x}\left( {x + 1} \right)dx = dt\] Substituting above values in given equation \[I = \int {\frac{{dt}}{{{{\cos }^2}t}}} \] \[I = \int {{{\sec }^2}tdt} \] I= tan(t) + c
I= tan(xex) + c (B) is correct option
Exercise 7.2 ⇐
⇒ Exercise 7.4