Mathematics

Evaluate the following:
$$\displaystyle \int_{0}^{\pi} xlog \,sin \,x \,dx$$


SOLUTION
Let $$I = \displaystyle \int_{0}^{\pi} xlog \,sin \,x \,dx$$....(i)
$$I = \displaystyle \int_{0}^{\pi} (\pi - x) log \,sin (\pi - x)dx$$
$$= \displaystyle \int_{0}^{\pi} (\pi -x) log \,sin \,x \,dx$$ ...(ii)
$$2I = \pi \displaystyle \int_{0}^{\pi} log \,sin \,x \,dx$$ .....(iii)
$$2I = 2\pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x \,dx [\because \displaystyle \int{-}^{2a} f(x) = 2 \displaystyle \int_{0}^{a} f(x) \,dx]$$
$$I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x \,dx$$ .....(iv)

Now, $$I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin (\pi / 2 - x)dx$$ ....(v)
On adding Eqs. (iv) and (v), we get
$$2I = \pi \displaystyle \int_{0}^{\pi/2} (log \,sin \,x + log \,cos \,x)dx$$
$$2I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x \,cos \,x \,dx$$
$$= \pi \displaystyle \int_{0}^{\pi / 22} log \dfrac{2 \,sin \,x \,cos \,x}{2}dx$$
$$2I = \pi \displaystyle \int_{0}^{\pi / 2} (log \,sin \,2x - log \,2)dx$$
$$2I = \pi \displaystyle \int_{0}^{\pi / 2} log \,sin \,2x dx - \pi \displaystyle \int_{0}^{\pi / 2} log \,2 \,dx$$
Put $$2x = t \Rightarrow dx = \dfrac{1}{2}dt$$
As $$x \rightarrow 0,$$ then $$t \rightarrow 0$$
and $$x \rightarrow \dfrac{\pi}{2},$$ then $$t \rightarrow \pi$$
$$\therefore 2I = \dfrac{\pi}{2} \displaystyle \int_{0}^{\pi} log \,sin \,t \,dt - \dfrac{\pi^2}{2} log \,2$$
$$\Rightarrow 2I = \dfrac{\pi}{2} \displaystyle \int_{0}^{\pi} log \,sin \,x \,dx - \dfrac{\pi^2}{2} log \,2$$
$$\Rightarrow 2I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x\, dx - \frac{\pi^2}{2} log \,2$$
$$\Rightarrow 2I = I - \frac{\pi^2}{2} log \,2$$ (from Eqn (iv))
$$\therefore I = -\dfrac{\pi^2}{2} log \,2 = \dfrac{\pi^2}{2} log \left ( \dfrac{1}{2} \right )$$
View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
$$\displaystyle \underset{0}{\overset{\pi}{\int}} \dfrac{x \, dx}{ 1 + \sin \, x}$$ =
  • A. $$\dfrac{\pi}{6} $$
  • B. $$\dfrac{\pi}{3}$$
  • C. Cannot be valued
  • D. $$\pi$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Hard
$$\displaystyle \int e^{2x} (\sqrt{3}cosx-sinx)dx=$$
  • A. $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx -(\sqrt{3}-2)sinx]+c$$
  • B. $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx+(\sqrt{3}-2)\cos x]+c$$
  • C. $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx-(\sqrt{3}-2)cosx]+c$$
  • D. $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx+(\sqrt{3}-2)sin x]+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Hard
Find $$\int \dfrac{(x^2+1) e^x}{(x+1)^2} dx$$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Medium
$$\displaystyle \int x\tan x\sec^{2}xdx=$$
  • A. $$\frac{1}{2}[x\tan^{2} x-tanx -x]+c$$
  • B. $$\displaystyle \frac{1}{2}[x\tan^{2}x+ tanx-x]+c$$
  • C. $$x\tan^{2} x-tanx+x+c$$
  • D. $$\displaystyle \frac{1}{2}[x\tan^{2} x-tanx +x]+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Easy
Evaluate $$\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer