Mathematics

# The value of $\displaystyle\int^{2\pi}_{0}\dfrac{x\sin^8x}{\sin^8x+\cos^8x}dx$ is equal to?

$\pi^2$

##### SOLUTION
Let $I = \displaystyle \int_0^{2 \pi} \dfrac{x \sin^8 x}{\sin^8 x + \cos^8 x}$ ___(1)

$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 (2\pi - x)}{\sin^8 (2 \pi - x) + \cos^8 (2 \pi - x)}$

$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 x}{\sin^8 x + \cos^8 x}$ ___(2)

Add (1) + (2) we get,

$2I = \displaystyle \int_0^{2 \pi} \dfrac{2 \pi \sin^8 x}{\sin^8 x + \cos^8 x}$

$2I = 2 \times 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x}$

$I = 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$

$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$ __(3)

$I = 4\pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 \left(\dfrac{\pi}{2} - x \right)}{\sin^8 \left(\dfrac{\pi}{2} - x \right) + \cos^8 \left(\dfrac{\pi}{2} - x \right)}$

$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\cos^8 (x)}{\sin^8 x + \cos^8 x}$ __(4)

add (3) + (4) we get

$I = 2 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x + \cos^8 x}{\sin^8 x + \cos^8 x} dx = 2\pi \displaystyle \int_0^{\pi/2} 1 dx = 2 \pi \times \dfrac{\pi}{2} = \pi^2$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Single Correct Hard
If $f(x) = x - x^2 +1$ & $g(x)=max\left \{ f(t) ;0\leq t< x \right \}$, then $\overset {1}{\underset { 0 }{ \int } } g (x) dx = ?$
• A. $\dfrac{7}{6}$
• B. $\dfrac{5}{4}$
• C. none of these
• D. $\dfrac{29}{24}$

1 Verified Answer | Published on 17th 09, 2020

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Solve $\int \:x^5\cdot \sqrt{a^3+x^3}dx$

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Q3 Subjective Hard
$\displaystyle\int _{ 0 }^{ 10 }{ \dfrac { { x }^{ 10 } }{ { \left( 10-x \right) }^{ 10 }+{ x }^{ 10 } } dx }$ is equal to

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Q4 Subjective Medium
Evaluate: $\displaystyle \int \cos^3 x \,\sin \,x \,dx$.

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Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$