Question

Evaluate the Integral

\(\int {cose{c^4}x\,co{t^6}x\,dx} \)

Short answer

The answer is \( - \frac{{co{t^9}x}}{9} - \frac{{co{t^7}x}}{7} + c\) where c is an integral constant.\(\)

First, convert\(cose{c^4}x\)in terms of\(cot\,x\)using trigonometric identities and then substituting\(cot\,x\) with\(u\). Solve the definite integral.

Step-by-step solution

Converting \(cose{c^4}x\) in \(cot\,x\) using trigonometric identity.

\(\begin{aligned}{l}\int {cose{c^4}x\,co{t^6}x\,dx} \\ &= \int {cose{c^2}x\,co{t^6}x\,cose{c^2}x\,dx} \\ &= \int {\left( {co{t^2}x + 1} \right)\,co{t^6}x\,cose{c^2}x\,dx} \end{aligned}\)

Putting \(cot\,x = u\).

Let \(cot\,x = u\)

\(\begin{aligned}{l} \Rightarrow - cose{c^2}x\,dx &= du\\ \Rightarrow - du &= cose{c^2}xdx\,\,....(1)\end{aligned}\)

So,

\(\begin{aligned}{l}\int {\left( {co{t^2}x + 1} \right)co{t^6}x} cose{c^2}xdx\\ &= \int {\left( {{u^2} + 1} \right)} {u^6}\left( { - du} \right)\,\,\,\,\left( {From\left( 1 \right)} \right)\\ &= - \int {\left( {{u^2} + 1} \right){u^6}\,du} \end{aligned}\)

Applying linearity and solving the integral.

\( = - \int {\left( {{u^2} + 1} \right){u^6}\,du} \)

\( = - \int {\left( {{u^8} + {u^6}} \right)\,du} \)

\( = - \int {{u^8}\,du} - \int {{u^6}\,du} \)

\( = - \frac{{{u^9}}}{9} - \frac{{{u^7}}}{7} + c\)

(where c is an integral constant)

Putting back \(u = cot\,x\).

\(\begin{aligned}{l} &= - \frac{{{u^9}}}{9} - \frac{{{u^7}}}{7} + c\\ &= - \frac{{cot{\,^9}x}}{9} - \frac{{co{t^7}x}}{7} + c\end{aligned}\)

Hence, \(\int {co{t^6}x} \,cose{c^4}x \,dx = - \frac{{cot{\,^9}x}}{9} - \frac{{co{t^7}x}}{7} + c\) where c is the integral constant.

\(\)\(\)