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Found in: Page 326

### Essential Calculus: Early Transcendentals

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

# $$\int {{{\cos }^2}} x{\tan ^3}xdx$$, Evaluate this integral

To evaluate this integral, we will use Trigonometric identities given as:

$$\tan x = \frac{{\sin x}}{{\cos x}}$$ and $${\sin ^2}x + {\cos ^2}x = 1$$

Firstly we will use $$\tan x = \frac{{\sin x}}{{\cos x}}$$ to change the integral in sine cosine functions as it will be easy to solve. Then we will use $${\sin ^2}x + {\cos ^2}x = 1$$ followed by substitution method.

$$\sqrt {{{\cos }^2}} x{\tan ^3}xdx = \frac{1}{2}{\cos ^2}x - cn|\cos x| + c$$

See the step by step solution

## Step 1 : Using general Trigonometric Identities

Let I = $$\int {{{\cos }^2}} x{\tan ^3}xdx$$

As, $$\tan x = \frac{{\sin x}}{{\cos x}}$$

$$= \int {{{\cos }^2}} x\frac{{{{\sin }^3}x}}{{{{\cos }^3}x}}dx$$

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

## Step 2 : Writing $${\sin ^3}x$$ as $${\sin ^2}x\sin x$$

As $${\sin ^2}x + {\cos ^2}x = 1$$

I = $$\int {\frac{{{{\sin }^2}x\sin x}}{{\cos x}}} dx$$

= $$\int {\frac{{(1 - {{\cos }^2}x)\sin x}}{{\cos x}}} dx$$

## Step 3: Substitution Method

Let u = $$\cos x$$ (=) du = $$- \sin xdx$$

(=) du = $$\sin xdx$$

\begin{aligned}{l} - \int {\frac{{1 - {u^2}}}{u}} dx\\ &= \int {\frac{{{u^2} - 1}}{u}} dx\\ &= \int {(u - \frac{1}{u}} )dx\end{aligned}

## Step 4: Splitting of Integral

Split this integral and use

$$\int {\frac{1}{x}} dx = ln|x|$$

\begin{aligned}{l}I &= \int {udu - \int {\frac{1}{u}} } du\\ &= \frac{{{u^2}}}{2} - ln|u|\end{aligned}

## Step 5 :Re-Substitution of the value of u

Now, Re-Substitute to get the answer

$$I = \frac{{{{\cos }^2}x}}{2} - ln|\cos x| + c$$

Hence the value of Integral

$$\int {{{\cos }^2}x{{\tan }^3}xdx = \frac{{{{\cos }^2}x}}{2}} - \ln |\cos x| + c$$

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