Mathematics

# Evaluate: $\displaystyle\int \dfrac { \cos ^ { 3 } x } { \sin ^ { 2 } x + \sin x }$

##### SOLUTION
$I=\displaystyle\int\dfrac{\cos^3x}{\sin^2x+\sin x}dx$

$I=\displaystyle\int\dfrac{(\cos^2x).\cos x}{\sin^2x+\sin x}dx$

$I=\displaystyle\int\dfrac{(1-\sin^2x).\cos x}{\sin^2x+\sin x}dx$

Let $u=\sin x$

$du=\cos xdx$

$\displaystyle=\int \dfrac{(1-u^2)\cos x}{u^2+u}.\dfrac{du}{\cos x}$

$\displaystyle=\int \dfrac{(1+u)(1-u)}{u(u+1)}du$

$\displaystyle=\int \dfrac{1}{u}du-\int \dfrac{u}{u}du$

$=ln|u|-u+c$

$=ln|\sin x|-\sin x+c$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int\log\sqrt{x+1}dx=$
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• B. $\displaystyle \frac{1}{2}[x \displaystyle \log(x+1)-\frac{x^{2}}{2}]+c$
• C. $\displaystyle \frac{1}{2}[(x+1)\displaystyle \log(x+1)-\frac{x}{2}]+c$
• D. $\displaystyle \frac{1}{2}[(x+1)\log (x+1)-x]+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
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Q3 Subjective Medium
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Q4 Subjective Medium
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