Mathematics

Evaluate: $$\displaystyle\int \sec ^{4}x. \mathrm{cosec} ^{2}xdx$$


ANSWER

$$\displaystyle \frac{1}{3}t^{3}+2t-\frac{1}{t}.$$


SOLUTION
$$\displaystyle I=\int \frac{1}{\cos ^{4}x\sin ^{2}x}dx=\int \frac{\left ( \sin ^{2}x+\cos ^{2}x \right )^{2}}{\cos ^{4}x\sin ^{2}x}dx$$ $$\displaystyle =\int \left [ \frac{\sin ^{2}x}{\cos ^{4}x}+\frac{1}{\sin ^{2}x} +\frac{2}{\cos ^{2}x}\right ]dx$$ $$\displaystyle =\int \left ( \sec ^{2}x\tan ^{2}x+co\sec ^{2}x+2\sec ^{2}x \right )dx=\frac{1}{3}\tan ^{3}x-\cot x+2\tan x$$ Alt. Divide above and below by $$\displaystyle \cos ^{4+2}x=\cos ^{6}x$$ $$\displaystyle \therefore =\int \frac{\sec ^{4}x,\sec ^{2}x}{\tan ^{2}x}dx=\int \frac{\left ( 1+\tan ^{2}x \right )^{2}}{\tan ^{2}x}\sec ^{2}xdx$$ $$\displaystyle =\int \frac{\left ( 1+t^{2} \right )^{2}}{t^{2}}dt=\int \left ( t^{2}+2+\frac{1}{t^{2}} \right )dt$$ $$\displaystyle =\frac{1}{3}t^{3}+2t-\frac{1}{t}etc.$$
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Single Correct Medium Published on 17th 09, 2020
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