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Essential Calculus: Early Transcendentals
Found in: Page 326
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

\(\int {{{\cos }^2}} x{\tan ^3}xdx\), 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 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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