Passage

Let $$\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$$

Then answer the following question.
Mathematics

$$\int_{0}^{\pi }xf\left ( x \right )dx=$$


ANSWER

$$\displaystyle \frac{8}{\pi ^{2}}$$


SOLUTION
Given $$\displaystyle =\int_{0}^{\pi }xf\left ( x \right )dx=\dfrac{1}{2}\int_{0}^{\pi }\dfrac{\left ( 2x-\pi +\pi  \right )\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }dx$$
   $$=\displaystyle \dfrac{1}{2}\int_{0}^{\pi }\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )dx+\dfrac{\pi }{2}\int_{0}^{\pi }f\left ( x \right )dx$$
   $$=\displaystyle \dfrac{1}{2}\int_{0}^{\pi }\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )dx+0$$          (As $$\int_{0}^{\pi }f\left ( x \right )dx=0$$)
   $$\displaystyle =\dfrac{2}{\pi }\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}\dfrac{2t}{\pi }\sin tdt=\dfrac{4}{\pi ^{2}}\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}t\sin tdt=\dfrac{8}{\pi ^{2}}\int_{0}^{\dfrac{\pi }{2}}t\sin tdt=\dfrac{8}{\pi ^{2}}$$
(Substituting $$\displaystyle \dfrac{\pi }{2}\cos x=t$$ then for $$x=0$$, $$\displaystyle t=\dfrac{\pi }{2}$$ and $$x=\pi $$, $$\displaystyle t=-\dfrac{\pi }{2}$$)
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Mathematics

$$\int_{0}^{\pi }f\left ( x \right )dx=$$


ANSWER

$$0$$


SOLUTION
$$\displaystyle I=\int_{0}^{\pi }f\left ( x \right )dx=\int_{0}^{\pi }\dfrac{\sin 2x\sin \left ( \dfrac{\pi }{2}.\cos x \right )}{2x-\pi }dx$$
$$\therefore $$   $$\displaystyle I=\int_{0}^{\pi }\dfrac{\sin 2\left ( \pi -x \right )\sin \left ( \dfrac{\pi }{2}.\cos \left ( \pi -x \right ) \right )}{2\left ( \pi -x \right )-\pi }dx$$
     $$\displaystyle =\int_{0}^{\pi }\dfrac{\left ( -1 \right )\sin 2x.\sin \left ( \dfrac{\pi }{2}\cos x \right )}{\pi -2x}dx=-I$$
$$\therefore $$   $$2I=0$$     $$\therefore $$   $$I=0$$
View Full Answer

Its FREE, you're just one step away


Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
Integrate the function 
$$x\sqrt {x+2}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 One Word Medium
$$\displaystyle \int_{0}^{\infty }\left ( \cot ^{-1}x \right )^{2}dx= \frac{\pi}{k} \log 2$$. Find the value of $$k$$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
If $$\displaystyle \int \frac{\log (t + \sqrt{1 + t^2})}{\sqrt{1 + t^2}} dt = \frac{1}{2} (g (t))^2 + C$$ where C is a constant, then $$g(2)$$ is equal to
  • A. $$\displaystyle \frac{1}{\sqrt 5} \log (2 + \sqrt 5)$$
  • B. $$2 \log (2 + \sqrt 5)$$
  • C. $$\displaystyle \frac{1}{2}\log (2 + \sqrt 5)$$
  • D. $$\log (2 + \sqrt 5)$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Medium
If $$\displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\tan \theta}{\sqrt{2k \sec \theta}}d\theta=1-\dfrac{1}{\sqrt{2}},(k>0),$$ then the value of k is :
  • A. $$\dfrac{1}{2}$$
  • B. $$4$$
  • C. $$1$$
  • D. $$2$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Medium
Consider two differentiable functions $$f(x), g(x)$$ satisfying $$\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$$ & $$\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$$. where $$\displaystyle f(x)>0    \forall  x \in  R$$

On the basis of above information, answer the following questions :

Asked in: Mathematics - Limits and Derivatives


1 Verified Answer | Published on 17th 08, 2020

View Answer