Question

Evaluate: $\int \sqrt{ \frac{\cos^{3}x}{\sin^{11}x}}\ dx$

Answer 1

Answer:

$\displaystyle\bf\int{\sqrt{\frac{cos^3x}{sin^{11}x}}}\,dx=-[\frac{2}{5}cot^{\frac{5}{2}}x+\frac{2}{9}cot^{\frac{9}{2}}x]+C$

Step-by-step explanation:

$\displaystyle\int{\sqrt{\frac{cos^3x}{sin^{11}x}}}\,dx$

$=\displaystyle\int{\frac{cosx}{sin^5x}\sqrt{\frac{cosx}{sinx}}}\,dx$

$=\displaystyle\int{\frac{cosx}{sinx}\frac{1}{sin^4x}\sqrt{\frac{cosx}{sinx}}}\,dx$

$=\displaystyle\int{cotx\,cosec^4x\,\sqrt{cotx}}\,dx$

$=\displaystyle\int{cot^{\frac{3}{2}}x\,cosec^2x\,cosec^2x}\,dx$

$=\displaystyle\int{cot^{\frac{3}{2}}x(1+cot^2x)\,cosec^2x}\,dx$

$=\displaystyle\int{[cot^{\frac{3}{2}}x+cot^{\frac{7}{2}}x]\,cosec^2x}\,dx$

$\boxed{\begin{minipage}{4cm} Take\,t=cotx\\ \\ \frac{dt}{dx}=-cosec^2x\\ \\ \implies\,-dt=cosec^2x\,dx \end{minipage}}$

$=\displaystyle\int{[t^{\frac{3}{2}}x+t^{\frac{7}{2}}]}(-dt)$

$=\displaystyle\,-[\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\frac{t^{\frac{9}{2}}}{\frac{9}{2}}]+C$

$=\displaystyle\,-[\frac{2}{5}t^{\frac{5}{2}}+\frac{2}{9}t^{\frac{9}{2}}]+C$

$=\displaystyle\,-[\frac{2}{5}cot^{\frac{5}{2}}x+\frac{2}{9}cot^{\frac{9}{2}}x]+C$

$\implies\displaystyle\bf\int{\sqrt{\frac{cos^3x}{sin^{11}x}}}\,dx=-[\frac{2}{5}cot^{\frac{5}{2}}x+\frac{2}{9}cot^{\frac{9}{2}}x]+C$

Answer 2

Answer:

$\mathbf{-\frac{2}{5}cot^{\frac{5}{2}}x+\frac{2}{9}cot^{\frac{9}{2}}x+C}$

Step-by-step explanation:

In this question,

We have to evaluate

$\mathbf{\displaystyle\int{\sqrt{\frac{cos^3x}{sin^{11}x}}}},dx$

Therefore, According to the question,

= $\displaystyle\int{\frac{cosx}{sin^5x}\sqrt{\frac{cosx}{sinx}}}\,dx$

= $\displaystyle\int{\frac{cosx}{sinx}\frac{1}{sin^4x}\sqrt{\frac{cosx}{sinx}}}\,dx$

= $\displaystyle\int{cotx\,cosec^4x\,\sqrt{cotx}}\,dx$

= $\displaystyle\int{cot^{\frac{3}{2}}x\,cosec^2x\,cosec^2x}\,dx$

= $\displaystyle\int{cot^{\frac{3}{2}}x(1+cot^2x)\,cosec^2x}\,dx$

=$\displaystyle\int{[cot^{\frac{3}{2}}x+cot^{\frac{7}{2}}x]\,cosec^2x}\,dx$

$\mathbf{\begin{minipage}{4cm} Take\,t=cotx\\ \\ \frac{dt}{dx}=-cosec^2x\\ \\ \implies\,-dt=cosec^2x\,dx \end{minipage}}$

=$\displaystyle\int{[t^{\frac{3}{2}}x+t^{\frac{7}{2}}]}(-dt)$

=$\displaystyle\,-[\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\frac{t^{\frac{9}{2}}}{\frac{9}{2}}]+C$

=$\displaystyle\,-[\frac{2}{5}t^{\frac{5}{2}}+\frac{2}{9}t^{\frac{9}{2}}]+C$

=$\displaystyle\,-[\frac{2}{5}cot^{\frac{5}{2}}x+\frac{2}{9}cot^{\frac{9}{2}}x]+C$

$\implies\displaystyle\bf\int{\sqrt{\frac{cos^3x}{sin^{11}x}}}\,dx=-[\frac{2}{5}cot^{\frac{5}{2}}x+\frac{2}{9}cot^{\frac{9}{2}}x]+C$