Question

$\int \frac{1}{{\sin }^{3}x{\cos }^{5}x}dx=$

• A. $3\log \tan x+\frac{3}{2}{\tan }^{2}x+\frac{1}{4}{\tan }^{4}x-\frac{1}{2}{\cot }^{2}x+c$
• B. $3\log \tan x+\frac{3}{2}{\tan }^{2}x+\frac{1}{4}{\tan }^{4}x-\frac{1}{2}{\tan }^{2}x+c$
• C. $3\log \cot x+\frac{3}{2}{\cot }^{2}x+\frac{1}{4}{\cot }^{4}x-\frac{1}{2}{\tan }^{2}x+c$
• D. $3\log \cot x+\frac{3}{2}{\cot }^{2}x+\frac{1}{4}{\cot }^{4}x-\frac{1}{2}{\cot }^{2}x+c$

Answer 1

The correct option is A $3\log \tan x+\frac{3}{2}{\tan }^{2}x+\frac{1}{4}{\tan }^{4}x-\frac{1}{2}{\cot }^{2}x+c$

$\int \frac{dx}{{\sin }^{3}x{\cos }^{5}x}$

$sinx=\frac{\tan x}{\sec x}$

$cosx=\frac{1}{\sec x}$

$=\int \frac{{\sec }^{8}xdx}{{\tan }^{3}x}$

$=\int \frac{{\sec }^{2}x{(1+{\tan }^{2}x)}^{3}dx}{{\tan }^{3}x}$

$\Rightarrow \tan x=t$

${sec}^{2}xdx=dt$

$=\int \frac{{(1+t^2)}^3}{t^3}dt=\int In(t^3+3t+\frac{3}{t}+\frac{1}{t^3})dt$

$=[\frac{t^4}{4}+\frac{{3t}^2}{2}+3Int-\frac{1}{2t^2}]+c$

$\Rightarrow \frac{{\left(\tan x\right)}^4}{4}+\frac{3{\left(\tan x\right)}^2}{2}+3In\left(\tan x\right)-\frac{{\cot }^{2}x}{2}+c$