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\[I=\int \frac{1}{\sin x{{\cos }^{2}}x}dx\]
\[I=\int \frac{1}{\sin x{{\cos }^{2}}x}dx\]
CBSE | Class 12 | Exercise 15A | Integration Using Partial Fractions | Maths | RS Aggarwal
\[I=\int \frac{1}{\sin x{{\cos }^{2}}x}dx\]

$=\int \left( \frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x{{\cos }^{2}}x} \right)dx$

$=\int \frac{{{\sin }^{2}}x}{\sin x{{\cos }^{2}}x}dx+\int \frac{{{\cos }^{2}}x}{\sin x{{\cos }^{2}}x}dx$

$=\int \frac{\sin x}{{{\cos }^{2}}x}dx+\int \frac{1}{\sin x}dx$

$=\int (\tan x\sec x+\cos ecx)dx$

$=\sec x-\frac{1}{2}\log {{\cot }^{2}}\frac{x}{2}=\sec x-\frac{1}{2}\log \left( \frac{1+\cos x}{1-\cos x} \right)+c$

i

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