Question

Find the integral of $\int co{t}^5\left(x\right)·\cos e{c}^2\left(x\right)dx$

Answer 1

Solve the given integral

Given: $\int co{t}^5\left(x\right)·\cos e{c}^2\left(x\right)dx$

Let $cot\left(x\right)=t$

Differentiate the above equation we get

$$\begin{gathered} \Rightarrow \frac{d}{dx}cot\left(x\right)=\frac{dt}{dx} \\ \Rightarrow -\cos e{c}^2\left(x\right)=\frac{dt}{dx} \\ \Rightarrow -\cos e{c}^2\left(x\right)dx=dt \end{gathered}$$

Now put the values of $cot\left(x\right),-\cos e{c}^2\left(x\right)dx$

$$\begin{gathered} =-\int t^{5}dt \\ =-\frac{t^6}{6}+C \\ =-\frac{1}{6}\left[co{t}^6\left(x\right)\right]+C\mathbf{}\left[\because t=\cot \left(x\right)\right] \end{gathered}$$

Hence the required answer is $-\frac{1}{6}co{t}^{6}x+C$