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Found in: Page 326

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the Integral$$\int {cose{c^4}x\,co{t^6}x\,dx}$$

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.

See the step by step solution

## Step 1: 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}

## Step 2: 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}

## Step 3: 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)

## Step 4: 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.