Mathematics

# Solve :$I=\int \sin^{6}{x}dx$

##### SOLUTION
$\displaystyle I = \int sin^{6}x\,dx$
$\displaystyle = \int (sin^{2}x)^{2}(sin^{2}x)dx$
$\displaystyle = \int (\frac{1-cos2x}{2})^{2}(\frac{1-cos2x}{2})dx$
$\displaystyle = \frac{1}{8}\int (1-2cos2x+cos^{2}x)(1-cos2x)dx$
$\displaystyle = \frac{1}{8}\int (1-2cos2x+\frac{1+cos4x}{2})(1-cos2x)dx$
$\displaystyle = \frac{1}{16}\int (2-4cos2x+cos4x)(1-cos2x)dx$
$\displaystyle = \frac{1}{16}\int (3-4cos2x+cos4x)(1-cos2x)dx$
$\displaystyle = \frac{1}{16}\int (3-4cos2x-3cos2x+cos4x+4cos^{2}2x-cos4xcos2x)dx$
$\displaystyle = \frac{1}{16}\int [3-7cos2x+cos4x+2\times 2cos^{2}2x-\frac{1}{2}.2cos^{4}xcos2x]dx$
$\displaystyle = \frac{1}{16}[3-7cos2x+cos^{4}x+2(1+cos^{4}x)-\frac{1}{2}(cos6x+cos2x)]dx$
$\displaystyle = \frac{1}{16}(3-7cos2x+cos^{4}x+2+2cos^{4}x-\frac{1}{2}(cos6x+cos2x))dx$
$\displaystyle = \frac{1}{32}\int [6-14cos2x+2cos4x+4+4cos4x-cos6x-cos2x]dx$
$\displaystyle = \frac{1}{32}[10-15cos2x+6cos4x-cos6x]dx$
$\displaystyle \frac{1}{32}[10-15cos2x+6cos4x-cos6x]dx$
$\displaystyle \frac{1}{32}[10x-\frac{15sin2x}{2}+\frac{6sin4x}{4}-\frac{sin6x}{6}]dx$
$\displaystyle \frac{1}{32}[10x-\frac{15sin2x}{2}+\frac{3}{2}sin4x]$
$\displaystyle \frac{1}{92}[60x-45sin2x+9sin4x-sin6x]+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
Evaluate $\displaystyle\int^{\pi/3}_{\pi/6}\dfrac{dx}{1+\sqrt{\tan x}}$.
• A. $\cfrac{7\pi}{12}$
• B. $\cfrac{5\pi}{12}$
• C. None of these
• D. $\cfrac{\pi}{12}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Let $\displaystyle \frac{df\left ( x \right )}{dx}=\frac{e^{\sin x}}{x}, x> 0$. If $\displaystyle \int_{1}^{4}\displaystyle \frac{3e^{\sin x^{3}}}{x}dx=f\left ( k \right )-f\left ( 1 \right )$ then one of the possible values of $k$ is
• A. $16$
• B. $63$
• C. $15$
• D. $64$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle\int_{-\pi}^{\pi}\sin^{3}x \cos^{2}x\ dx$ is equal to
• A. $1$
• B. $2$
• C. $3$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate: $\displaystyle \int \sqrt[9]{x^{-8}}dx$
• A. $9x^{\frac{8}{9}}+c$
• B. $x^{\frac{1}{9}}+c$
• C. $x^{\frac{8}{9}}+c$
• D. $9x^{\frac{1}{9}}+c$

Prove that $\displaystyle\int_0^{\pi/2}$ $ln(\sin x)dx=\displaystyle\int_0^{\pi/2}ln(cos x)dx=\int_0^{\pi/2}\,\,ln(sin2x)dx=-\dfrac{\pi}{2}.ln 2$.