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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals3 Marks
Question
Find the integrals of the function $\frac{1}{\sin x \cos ^{3} x}$
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Answer

Clearly, $\frac{1}{\sin x \cos ^{3} x}=\frac{\sin x}{\cos ^{3} x}+\frac{1}{\sin x \cos x}$ 
$= \tan x \sec ^{2} x+\frac{\frac{1}{\cos ^{2} x}}{\frac{\sin x \cos x}{\cos ^{2} x}}$ 
$= \tan x \sec ^{2} x+\frac{\sec ^{2} x}{\tan x}$ 
Now, $\int \frac{1}{\sin x \cos ^{3} x} d x=\int \tan x \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x$ 
Let tan x = t
$\Rightarrow$ sec2 x dx = dt
$\Rightarrow \int \frac{1}{\sin x \cos ^{3} x} d x=\int \operatorname{td} t+\int \frac{1}{t} d t$ 
$= \frac{t^{2}}{2}+\log |t|+C$ 
$= \frac{1}{2} \tan ^{2} x+\log |\tan x|+C$
 

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