CBSE Class 12 Mathematics Integration Worksheet Set 03

CBSE Class 12 Mathematics Integration Worksheet (10). CBSE issues sample papers every year for students for class 12 board exams. Students should solve the CBSE issued sample papers to understand the pattern of the question paper which will come in class 12 board exams this year. The sample papers have been provided with marking scheme. It’s always recommended to practice as many CBSE sample papers as possible before the board examinations. Sample papers should be always practiced in examination condition at home or school and the student should show the answers to teachers for checking or compare with the answers provided. Students can download the sample papers in pdf format free and score better marks in examinations. Refer to other links too for latest sample papers.

Type: Integration By Parts

Question. (a) \( I = \int x^2 \sin x \, dx \)      (b) \( I = \int x \sin^2 x \, dx \)
Answer:
(a) \( I = \int x^2 \sin x \, dx \)
Using Integration by Parts:
\( = x^2(-\cos x) - \int 2x \cdot (-\cos x) \, dx \)
\( = -x^2 \cos x + 2 \int x \cos x \, dx \)
\( = -x^2 \cos x + 2 \left[ x(\sin x) - \int (1) \cdot \sin x \, dx \right] \)
\( = -x^2 \cos x + 2 [x \sin x + \cos x] + c \)

(b) \( I = \int x \sin^2 x \, dx \)
\( = \int x \left( \frac{1 - \cos(2x)}{2} \right) \, dx \)
\( = \frac{1}{2} \int [x - x \cos(2x)] \, dx \)
\( = \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos(2x) \, dx \)
\( = \frac{1}{2} \cdot \frac{x^2}{2} - \frac{1}{2} \left[ x \frac{\sin(2x)}{2} - \int (1) \cdot \frac{\sin(2x)}{2} \, dx \right] \)
\( = \frac{x^2}{4} - \frac{1}{2} \left[ \frac{1}{2} x \sin(2x) - \frac{1}{2} \int \sin(2x) \, dx \right] \)
\( = \frac{x^2}{4} - \frac{1}{2} \left[ \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) \right] + c \)

Question. (a) \( I = \int \log x \, dx \)      (b) \( I = \int (\log x)^2 \, dx \)      (c) \( I = \int \frac{\log x}{x^2} \, dx \)
Answer:
(a) \( I = \int \log x \, dx \)
\( = \int \log x \cdot 1 \, dx \)
\( = \log x \cdot (x) - \int \frac{1}{x} \cdot (x) \, dx \)
\( = x \log x - \int 1 \, dx \)
\( = x \log x - x + c \)

(b) \( I = \int (\log x)^2 \, dx \)
\( = \int (\log x)^2 \cdot 1 \, dx \)
\( = (\log x)^2 \cdot x - \int \frac{2 \log x}{x} \cdot x \, dx \)
\( = x(\log x)^2 - 2 \int \log x \cdot 1 \, dx \)
\( = x(\log x)^2 - 2 \left[ \log x \cdot (x) - \int \frac{1}{x} \cdot x \, dx \right] \)
\( = x(\log x)^2 - 2 [x \log x - x] + c \)

(c) \( I = \int \frac{\log x}{x^2} \, dx \)
\( = \int \frac{1}{x^2} \cdot \log x \, dx \)
\( = \log x \left( -\frac{1}{x} \right) - \int \frac{1}{x} \left( -\frac{1}{x} \right) \, dx \)
\( = -\frac{1}{x} \log x + \int \frac{1}{x^2} \, dx \)
\( = -\frac{1}{x} \log x - \frac{1}{x} + c \)

Question. (a) \( I = \int \sin^{-1} x \, dx \)      (b) \( I = \int (\sin^{-1} x)^2 \, dx \)
Answer:
(a) \( I = \int \sin^{-1} x \, dx \)
\( = \int \sin^{-1} x \cdot 1 \, dx \)
\( = \sin^{-1} x \cdot (x) - \int \frac{1}{\sqrt{1-x^2}} \cdot x \, dx \)
Put \( 1 - x^2 = t \implies -2x \, dx = dt \implies x \, dx = -\frac{dt}{2} \)
\( \therefore I = x \sin^{-1} x + \frac{1}{2} \int \frac{dt}{\sqrt{t}} \)
\( = x \sin^{-1} x + \frac{1}{2} \times 2\sqrt{t} + c \)
\( = x \sin^{-1} x + \sqrt{1-x^2} + c \)

(b) \( I = \int (\sin^{-1} x)^2 \, dx \)
\( = \int (\sin^{-1} x)^2 \cdot 1 \, dx \)
\( = (\sin^{-1} x)^2 \cdot x - \int 2 \frac{\sin^{-1} x}{\sqrt{1-x^2}} \cdot x \, dx \)
\( = x(\sin^{-1} x)^2 - 2 \int x \frac{\sin^{-1} x}{\sqrt{1-x^2}} \, dx \)
Put \( \sin^{-1} x = t \implies x = \sin t \) and \( \frac{1}{\sqrt{1-x^2}} \, dx = dt \)
\( \dots I = x(\sin^{-1} x)^2 - 2 \int \sin t \cdot t \, dt \)
\( = x(\sin^{-1} x)^2 - 2 \left[ t(-\cos t) - \int (1)(-\cos t) \, dt \right] \)
\( = x(\sin^{-1} x)^2 - 2 [ -t \cos t + \sin t ] + c \)
\( = x(\sin^{-1} x)^2 - 2 \left[ -\sin^{-1} x \sqrt{1-x^2} + x \right] + c \)
\( \left\{ \text{since } \sin t = x \text{ then } \cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2} \right\} \)

Question. (a) \( I = \int \frac{x \tan^{-1} x}{(1+x^2)^{3/2}} \, dx \)      (b) \( I = \int \frac{x^2 \sin^{-1} x}{(1-x^2)^{3/2}} \, dx \)
Answer:
(a) \( I = \int \frac{x \tan^{-1} x}{(1+x^2)^{3/2}} \, dx \)
\( = \int \frac{x \tan^{-1} x}{\sqrt{1+x^2}(1+x^2)} \, dx \)
Put \( \tan^{-1} x = t \implies \frac{1}{1+x^2} \, dx = dt \)
also \( x = \tan t \)
\( \therefore I = \int \frac{xt}{\sqrt{1+x^2}} \, dt \)
put \( x = \tan t \)
\( I = \int \frac{\tan t \cdot t}{\sqrt{1+\tan^2 t}} \, dt \)
\( = \int t \frac{\tan t}{\sec t} \, dt \)
\( = \int t \sin t \, dt \)
\( = t(-\cos t) - \int (1)(-\cos t) \, dt \)
\( = -t \cos t + \sin t + c \)
\( \left[ \text{since } \tan t = x \implies P = x, B = 1 \implies H = \sqrt{x^2+1} \implies \cos t = \frac{B}{H} = \frac{1}{\sqrt{x^2+1}}, \sin t = \frac{P}{H} = \frac{x}{\sqrt{x^2+1}} \right] \)
\( = -\tan^{-1} x \cdot \frac{1}{\sqrt{x^2+1}} + \frac{x}{\sqrt{1+x^2}} + c \)

(b) \( I = \int \frac{x^2 \sin^{-1} x}{(1-x^2)^{3/2}} \, dx \)
\( = \int \frac{x^2 \sin^{-1} x}{(1-x^2)\sqrt{1-x^2}} \, dx \)
put \( \sin^{-1} x = t \implies (\sin t = x) \)
\( \therefore \frac{1}{\sqrt{1-x^2}} \, dx = dt \)
\( \therefore I = \int \frac{x^2 t}{(1-x^2)} \, dt \)
put \( \sin t = x \)
\( I = \int \frac{\sin^2 t \cdot t}{1 - \sin^2 t} \, dt \)
\( = \int \tan^2 t \cdot t \, dt \)
\( = \int (\sec^2 t - 1)t \, dt \)
\( = \int t \sec^2 t \, dt - \int t \, dt \)
\( = t(\tan t) - \int (1) \cdot \tan t \, dt - \frac{t^2}{2} + c \)
\( \left[ \text{since } \sin t = x \implies P = x, H = 1 \implies B = \sqrt{1-x^2} \implies \tan t = \frac{P}{B} = \frac{x}{\sqrt{1-x^2}}, \sec t = \frac{H}{B} = \frac{1}{\sqrt{1-x^2}} \right] \)
\( = t \tan t - \log | \sec t | - \frac{t^2}{2} + c \)
\( = \sin^{-1} x \cdot \frac{x}{\sqrt{1-x^2}} - \log \left| \frac{1}{\sqrt{1-x^2}} \right| - \frac{(\sin^{-1} x)^2}{2} + c \)
\( = \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \frac{1}{2} \log | 1 - x^2 | - \frac{(\sin^{-1} x)^2}{2} + c \)

Question. \( I = \int \frac{x - \sin x}{1 - \cos x} \, dx \)
Answer:
\( I = \int \frac{x - \sin x}{1 - \cos x} \, dx \)
\( = \int \frac{x - 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \, dx \)
Separate:
\( = \frac{1}{2} \int x \csc^2 \left( \frac{x}{2} \right) \, dx - \int \cot \left( \frac{x}{2} \right) \, dx \)
\( = \frac{1}{2} \left[ x \left( -2 \cot \left( \frac{x}{2} \right) \right) - \int 2 \left( -cot \left( \frac{x}{2} \right) \, dx \right) \right] - 2 \log \left| \sin \left( \frac{x}{2} \right) \right| \)
\( = \frac{1}{2} \left[ -2x \cot \left( \frac{x}{2} \right) + 2 \int \cot \left( \frac{x}{2} \right) \, dx \right] - 2 \log \left| \sin \frac{x}{2} \right| \)
\( = \frac{1}{2} \left[ -2x \cot \left( \frac{x}{2} \right) + 2 \times 2 \log \left| \sin \left( \frac{x}{2} \right) \right| \right] - 2 \log \left| \sin \frac{x}{2} \right| \)
\( = -x \cot \left( \frac{x}{2} \right) + 2 \log \left| \sin \left( \frac{x}{2} \right) \right| - 2 \log \left| \sin \left( \frac{x}{2} \right) \right| + c \)
\( = -x \cot \left( \frac{x}{2} \right) + c \)

Question. (a) \( I = \int x \sin^{-1} x \, dx \)      (b) \( I = \int \sin^{-1}\sqrt{x} \, dx \)
Answer:
(a) \( I = \int x \sin^{-1} x \, dx \)
\( = \sin^{-1} x \cdot \frac{x^2}{2} - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx \)
\( = \frac{x^2}{2} \sin^{-1} x + \frac{1}{2} \int \frac{-x^2}{\sqrt{1-x^2}} \, dx \)
\( = \frac{x^2}{2} \sin^{-1} x + \frac{1}{2} \int \frac{1-x^2-1}{\sqrt{1-x^2}} \, dx \)
\( = \frac{x^2}{2} \sin^{-1} x + \frac{1}{2} \int \sqrt{1-x^2} - \frac{1}{\sqrt{1-x^2}} \, dx \)
\( = \frac{x^2}{2} \sin^{-1} x + \frac{1}{2} \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) - \sin^{-1}(x) \right] + c \)
\( \left[ \text{since } \int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right] \)
\( = \frac{x^2}{2} \sin^{-1} x + \frac{1}{2} \left[ \frac{x}{2}\sqrt{1-x^2} - \frac{1}{2}\sin^{-1} x \right] + c \)

(b) \( I = \int \sin^{-1}\sqrt{x} \, dx \)
put \( \sqrt{x} = t \)
\( x = t^2 \)
\( dx = 2t \, dt \)
\( \therefore I = 2 \int \sin^{-1} t \cdot t \, dt \)
Proceed as above Ques.
\( = x \sin^{-1} x + \frac{1}{2} \sqrt{x - x^2} - \frac{1}{2} \sin^{-1}\sqrt{x} + c \)

Question. (a) \( I = \int \frac{\sin^{-1} x}{x^2} \, dx \)      (b) \( I = \int \frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}} \, dx \)
Answer:
(a) \( I = \int \frac{\sin^{-1} x}{x^2} \, dx \)
put \( \sin^{-1} x = t \)
\( x = \sin t \)
\( \therefore dx = \cos t \, dt \)
\( \therefore I = \int \frac{t \cos t}{\sin^2 t} \, dt \)
\( = \int t \csc t \cdot \cot t \, dt \)
\( = t(-\csc t) - \int (1)(-\csc t) \, dt \)
\( = \frac{-t}{\sin t} + \int \csc t \, dt \)
\( = \frac{-t}{\sin t} + \log | \csc t - \cot t | + c \)
\( = \frac{-\sin^{-1} x}{x} + \log \left| \frac{1}{\sin t} - \frac{\cot t}{\sin t} \right| + c \)
\( = \frac{-\sin^{-1} x}{x} + \log \left| \frac{1}{x} - \frac{\sqrt{1-x^2}}{x} \right| + c \)

(b) \( I = \int \frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}} \, dx \)
Property : \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \implies \sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x} = \frac{\pi}{2} \)
\( I = \int \frac{\sin^{-1}\sqrt{x} - \left( \frac{\pi}{2} - \sin^{-1}\sqrt{x} \right)}{\frac{\pi}{2}} \, dx \)
\( = \frac{2}{\pi} \int 2\sin^{-1}\sqrt{x} - \frac{\pi}{2} \, dx \)
\( = \frac{4}{\pi} \int \sin^{-1}\sqrt{x} \, dx - \int 1 \, dx \)
put \( \sqrt{x} = t \)
\( x = t^2 \)
\( dx = 2t \, dt \)
\( \therefore I = \frac{8}{\pi} \int \sin^{-1} t \cdot t \, dt - x \)
\( = \frac{8}{\pi} \left[ \sin^{-1} t \cdot \frac{t^2}{2} - \frac{1}{2} \int \frac{1}{\sqrt{1-t^2}} \cdot t^2 \, dt \right] - x \)
\( = \frac{8}{\pi} \left[ \frac{t^2}{2} \cdot \sin^{-1} t + \frac{1}{2} \int \frac{-t^2}{\sqrt{1-t^2}} \, dt \right] - x \)
\( = \frac{8}{\pi} \left[ \frac{t^2}{2} \sin^{-1} t + \frac{1}{2} \int \frac{1-t^2-1}{\sqrt{1-t^2}} \, dt \right] - x \)
\( = \frac{8}{\pi} \left[ \frac{t^2}{2} \sin^{-1} t + \frac{1}{2} \int \sqrt{1-t^2} - \frac{1}{\sqrt{1-t^2}} \, dt \right] - x \)
\( = \frac{8}{\pi} \left[ \frac{t^2}{2} \sin^{-1} t + \frac{1}{2} \left\{ \frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin^{-1}(t) - \sin^{-1}(t) \right\} \right] - x + c \)
\( = \frac{8}{\pi} \left[ \frac{t^2}{2} \sin^{-1} t + \frac{1}{2} \left\{ \frac{t}{2}\sqrt{1-t^2} - \frac{1}{2}\sin^{-1} t \right\} \right] - x + c \)
\( = \frac{8}{\pi} \left[ \frac{t^2}{2} \sin^{-1} t + \frac{t}{4}\sqrt{1+t^2} - \frac{1}{4}\sin^{-1} t \right] - x + c \)
replace t by \( \sqrt{x} \)
\( = \frac{8}{\pi} \left[ \frac{x}{2} \sin^{-1}\sqrt{x} + \frac{\sqrt{x}\sqrt{1-x}}{4} - \frac{\sin^{-1}\sqrt{x}}{4} \right] - x + c \)

Question. (a) \( I = \int \frac{\sqrt{x^2+1}[\log(x^2+1) - 2\log x]}{x^4} \, dx \)      (b) \( I = \int \tan^{-1}\sqrt{\frac{1-x}{1+x}} \, dx \)
Answer:
(a) \( I = \int \frac{\sqrt{x^2+1}[\log(x^2+1) - 2\log x]}{x^4} \, dx \)
\( = \int \frac{\sqrt{x^2+1}(\log(x^2+1) - \log(x^2))}{x^4} \, dx \)
\( = \int \frac{\sqrt{x^2+1}\left[ \log \left( \frac{x^2+1}{x^2} \right) \right]}{x^4} \, dx \)
take \( x^2 \) common:
\( = \int \frac{x\sqrt{1+\frac{1}{x^2}} \cdot \log\left(1+\frac{1}{x^2}\right)}{x^4} \, dx \)
\( = \int \frac{\sqrt{1+\frac{1}{x^2}} \cdot \log\left(1+\frac{1}{x^2}\right)}{x^3} \, dx \)
put \( 1 + \frac{1}{x^2} = t \)
\( -\frac{2}{x^3} \, dx = dt \implies \frac{1}{x^3} \, dx = \frac{-dt}{2} \)
\( \therefore I = \frac{-1}{2} \int \sqrt{t} \cdot \log t \, dt \)
\( = \frac{-1}{2} \left[ \log t \cdot \frac{2}{3}(t)^{3/2} - \frac{2}{3} \int \frac{1}{t} \cdot t^{3/2} \, dt \right] \)
\( = \frac{-1}{2} \left[ \frac{2}{3} \log t \cdot t^{3/2} - \frac{2}{3} \int t^{1/2} \, dt \right] \)
\( = \frac{-1}{2} \left[ \frac{2}{3} \log t \cdot t^{3/2} - 4t^{3/2} \right] + c \)
\( = \frac{-1}{2} \times \frac{2}{3} \times t^{3/2} \left[ \log t - \frac{2}{3} \right] + c  /> replacing t:
\( = \frac{-1}{3} \left( 1 + \frac{1}{x^2} \right)^{3/2} \cdot \left[ \log \left( 1 + \frac{1}{x^2} \right) - \frac{2}{3} \right] + c \)

(b) \( I = \int \tan^{-1}\sqrt{\frac{1-x}{1+x}} \, dx \)
put \( x = \cos(2\theta) \)
\( \therefore I = \int \tan^{-1}\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} \, dx \)
\( = \int \tan^{-1}\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} \, dx \)
\( = \int \tan^{-1}(\tan\theta) \, dx \)
\( = \int \theta \, dx \)
replacing \( \theta \):
\( = \frac{1}{2} \int \cos^{-1} x \, dx \)
\( = \frac{1}{2} \int \cos^{-1} x \cdot 1 \, dx \)
\( = \frac{1}{2} \left[ \cos^{-1} x \cdot x - \int \frac{-1}{\sqrt{1-x^2}} \cdot x \, dx \right] \)
\( = \frac{1}{2} \left[ x\cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} \, dx \right] \)
put \( 1 - x^2 = t \implies x \, dx = \frac{-dt}{2} \)
\( \therefore I = \frac{1}{2} \left[ x\cos^{-1} x - \frac{1}{2} \int \frac{dt}{\sqrt{t}} \right] \)
\( = \frac{1}{2} \left[ x\cos^{-1} x - \frac{1}{2} \times 2\sqrt{t} \right] + c \)
\( = \frac{1}{2} \left[ x\cos^{-1} x - \sqrt{1-x^2} \right] + c \)

Question. (a) \( I = \int \sin^{-1}\sqrt{\frac{x}{a+x}} \, dx \)      (b) \( I = \int \frac{x^2}{(x \sin x + \cos x)^2} \, dx \)
Answer:
(a) \( I = \int \sin^{-1}\sqrt{\frac{x}{a+x}} \, dx \)
put \( x = a \tan^2\theta \)
\( dx = 2a \tan\theta \cdot \sec^2\theta \, d\theta \)
\( \therefore I = 2a \int \sin^{-1}\sqrt{\frac{a \tan^2\theta}{a + a \tan^2\theta}} \cdot \tan\theta \cdot \sec^2\theta \, d\theta \)
\( = 2a \int \sin^{-1}\sqrt{\frac{\tan^2\theta}{\sec^2\theta}} \cdot \tan\theta \cdot \sec^2\theta \, d\theta \)
\( = 2a \int \sin^{-1}\sqrt{\sin^2\theta} \cdot \tan\theta \cdot \sec^2\theta \, d\theta \)
\( = 2a \int \sin^{-1}(\sin\theta) \cdot \tan\theta \cdot \sec^2\theta \, d\theta \)
\( = 2a \int \theta \cdot \tan\theta \cdot \sec^2\theta \, d\theta \)
\( = 2a \left[ \theta \cdot \int \tan\theta \sec^2\theta \, d\theta - \int \left( 1 \cdot \int \tan\theta \cdot \sec^2\theta \, d\theta \right) d\theta \right] \)
put \( \tan\theta = t \) in both integrals
\( \therefore \sec^2\theta \, d\theta = dt \)
\( = 2a \left[ \theta \cdot \int t \, dt - \int \left( \int t \, dt \right) d\theta \right] \)
\( = 2a \left[ \theta \cdot \frac{t^2}{2} - \int \left( \frac{t^2}{2} \right) d\theta \right] \)
replacing t:
\( = 2a \left[ \theta \cdot \frac{\tan^2\theta}{2} - \frac{1}{2} \int \tan^2\theta \, d\theta \right] \)
\( = 2a \left[ \theta \cdot \frac{\tan^2\theta}{2} - \frac{1}{2} \int (\sec^2\theta - 1) \, d\theta \right] \)
\( = 2a \left[ \theta \cdot \frac{\tan^2\theta}{2} - \frac{1}{2} \{ \tan\theta - \theta \} \right] + c \)
replacing \( \theta \) by \( \tan^{-1}\sqrt{\frac{x}{a}} \)
\( = a \left[ \tan^{-1}\sqrt{\frac{x}{a}} \cdot \frac{x}{a} - \left\{ \sqrt{\frac{x}{a}} - \tan^{-1}\sqrt{\frac{x}{a}} \right\} \right] + c \)
\( = a \left[ \frac{x}{a} \tan^{-1}\sqrt{\frac{x}{a}} - \sqrt{\frac{x}{a}} + \tan^{-1}\sqrt{\frac{x}{a}} \right] + c \)
\( = a \left[ \tan^{-1}\sqrt{\frac{x}{a}} \left( \frac{x}{a} + 1 \right) - \sqrt{\frac{x}{a}} \right] + c \)

(b) \( I = \int \frac{x^2}{(x \sin x + \cos x)^2} \, dx \)
adjustment:
\( = \int \frac{x \cdot x \cdot \cos x \cdot \sec x}{(x \sin x + \cos x)^2} \, dx \)
\( = (x \sec x) \cdot \int \frac{x \cos x}{(x \sin x + \cos x)^2} \, dx \)
Both parts:
\( = x \sec x \cdot \int \frac{x \cos x}{(x\sin x+\cos x)^2} - \int \frac{d}{dx}(x \sec x) \cdot \int \frac{x \cos x}{(x \sin x + \cos x)^2} \, dx \)
put \( x \sin x + \cos x = t \) in both integrals
\( (x\cos x + \sin x - \sin x) \, dx = dt \)
\( x \cos x \, dx = dt \)
\( \therefore I = x \sec x \int \frac{dt}{t^2} - \int (x \sec x \tan x + \sec x) \cdot \int \frac{dt}{t^2} \)
\( = x \sec x \left[ \frac{-1}{t} \right] - \int (x \sec x \tan x + \sec x) \left( \frac{-1}{t} \right) \, dx \)
replacing t by \( x \sin x + \cos x = t \)
\( = \frac{-x \sec x}{x\sin x+\cos x} - \int (x \sec x \tan x + \sec x) \left( \frac{-1}{x\sin x+\cos x} \right) \, dx \)
\( = \frac{-x \sec x}{x\sin x+\cos x} + \int \sec x (x \tan x + 1) \frac{1}{x\sin x+\cos x} \, dx \)
\( = \frac{-x \sec x}{x \sin x+\cos x} + \int \sec x \left( \frac{x \sin x+\cos x}{\cos x} \right) \left( \frac{1}{x\sin x+\cos x} \right) \, dx \)
\( = \frac{-x \sec x}{x \sin x+\cos x} - \int \sec^2 x \, dx \)
\( = \frac{-x \sec x}{x\sin x+\cos x} + \tan x + c \)

Question. \( I = \int \sec^3 x \, dx \)
Answer:
\( I = \int \sec^3 x \, dx \)
\( = \int \sec x \cdot \sec^2 x \, dx \)
\( = \sec x \cdot \tan x - \int \sec x \cdot \tan x \cdot \tan x \, dx \)
\( = \sec x \cdot \tan x - \int \sec x (\sec^2 x - 1) \, dx \)
\( = \sec x \cdot \tan x - \int \sec^3 x - \sec x \, dx \)
\( = \sec x \cdot \tan x - \int \sec^3 x \, dx + \int \sec x \, dx \)
\( I = \sec x \cdot \tan x - I + \log | \sec x + \tan x | \)
\( 2I = \sec x \tan x + \log | \sec x + \tan x | + c \)
\( \therefore I = \frac{1}{2} [\sec x \tan x + \log | \sec x + \tan x |] + c \)

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