Mathematics

# The value of $\displaystyle\int _{ 0 }^{ \pi /2 }{ \sin ^{ 6 }{ x } } dx$ is

$\cfrac{5\pi}{32}$

##### SOLUTION
$\displaystyle \int _{0}^{\pi /2} \sin ^6 x\ dx$

$I=\displaystyle \int _{0}^{\pi /2} \cos ^6 x\ dx$

$2I=\displaystyle \int _{0}^{\pi /2} (\sin ^6 x+\cos^6 x) dx$

$2I=\displaystyle \int _{0}^{\pi /2} (\sin^2 x)^3+(\cos^2 x)^3 dx =\displaystyle \int _{0}^{\pi /2} (\sin^2 x+\cos^2x)(\sin^4 x+\cos^4x-\sin^2\cos^2x) dx$

$2I=\displaystyle \int _{0}^{\pi /2} (\sin^2 x+\cos^2x)^2-(3\sin^2 x\cos^2x) dx$

$2I=\displaystyle \int _{0}^{\pi /2} (1-3\sin^2 x\cos^2 x)dx\quad \cos 2x=2\cos^2x-1 =1-2\sin^2x$

$2I=\displaystyle \int _{0}^{\pi /2} \left (1-\dfrac {3}{4} \sin^2x \right)dx$

$2I=\displaystyle \int _{0}^{\pi /2} \left (1-\dfrac {3}{4} \left (1-\dfrac {\cos 4x}{2}\right)\right)=\displaystyle \int _{0}^{\pi /2} \left (1-\dfrac {3}{8} +\dfrac {3}{8} \cos x\right)dx$

$2I=\left [\dfrac {5}{8}x+\dfrac {3}{8} \dfrac {\sin 4x}{4}\right]_0^{\pi /2}=\dfrac {5}{7}\dfrac {\pi}{2}$

$2I=\dfrac {5\pi}{16}\ \Rightarrow \ I=\dfrac {5\pi}{32}$

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

#### Realted Questions

Q1 Single Correct Medium
IF $f(x)=x^{2}$ for $0 \le x \le 1,\sqrt {x}$ for $1 \le x \le 2$ then $\int _{0}^{2}f(x)dx=$
• A. $\dfrac {4\sqrt {2}}{3}$
• B. $\dfrac {\sqrt {2}}{3}$
• C. $\dfrac {1}{3}$
• D. $\dfrac {4\sqrt {2}-1}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Solve $\displaystyle\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 2 }-5x+6 } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate:$\displaystyle\int_{0}^{\frac{\pi}{4}}{\ln{\left(1+\tan{x}\right)}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\int _{ -\pi }^{ \pi }{ \frac { 2x(1+sinx) }{ { 1+cos }^{ 2 }x } dx }$
• A. $\pi ^ 2$
• B. $zero$
• C. $\frac \pi 4$
• D. $\frac { { \pi }^{ 2 } }{ 4 }$

$\int {\dfrac{x}{{{x^4} - {x^2} + 1}}dx}$