Mathematics

$$\displaystyle \overset{\pi/2}{\underset{0}{\int}}{ \dfrac { \sin ^6x }{ \cos ^{ 6 }{ x } +\sin ^{ 6 }{ x }  }  }dx$$ is equal to:


ANSWER

$$\dfrac{\pi}{4}$$


SOLUTION
Let, $$I=$$$$\displaystyle \overset{\pi/2}{\underset{0}{\int}}{ \dfrac { \sin ^6x }{ \cos ^{ 6 }{ x } +\sin ^{ 6 }{ x }  }  }dx$$.......(1).
or, $$I=$$$$\displaystyle \overset{\pi/2}{\underset{0}{\int}}{ \dfrac { \cos ^6x }{ \sin ^{ 6 }{ x } +\cos ^{ 6 }{ x }  }  }dx$$ [ Using property of definite integral].......(2).
Now adding (1) and (2) we get,
$$2I=\displaystyle\int\limits_{0}^{\dfrac{\pi}{2}} \ dx$$
or, $$I=\dfrac{\pi}{4}$$.
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
Solve $$\int_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x}\cos xdx$$
  • A. $$\dfrac{2}{3}$$
  • B. $$\dfrac{5}{3}$$
  • C. $$-\dfrac{1}{3}$$
  • D. $$\dfrac{1}{3}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Hard
Evalaute the integral
$$\displaystyle \int_{0}^{1}\sin^{-1}(\frac{2x}{1+x^{2}})dx$$
  • A. $$\displaystyle \frac{\pi}{4}-\log 2$$
  • B. $$\displaystyle \frac{\pi}{2}+\log 2$$
  • C. $$\dfrac{\pi}{4}+\log 2$$
  • D. $$\displaystyle \frac{\pi}{2}-\log 2$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
$$\int { \sqrt { secx-1 }  } dx$$ is equal to 
  • A. $$2\ell n(cos\frac { x }{ 2 } +\sqrt { { cos }^{ 2 }\frac { x }{ 2 } -\frac { 1 }{ 2 } ) } +C$$
  • B. $$\ell n(cos\frac { x }{ 2 } +\sqrt { { cos }^{ 2 }\frac { x }{ 2 } -\frac { 1 }{ 2 } ) } +C$$
  • C. $$none$$ $$of$$ $$these$$
  • D. $$-2\ell n(cos\frac { x }{ 2 } +\sqrt { { cos }^{ 2 }\frac { x }{ 2 } -\frac { 1 }{ 2 } ) } +C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Medium
Find its:-
$$\mathop {Lt}\limits_{x \to \infty } \left( {\frac{{1 + {2^4} + {3^4} + ......{n^4}}}{{{n^5}}}} \right) - \mathop {Lt}\limits_{x \to \infty } \left( {\frac{{1 + {2^3} + {3^3} + ......{n^3}}}{{{n^5}}}} \right) = $$
  • A. $$0$$
  • B. $$\frac{1}{5}$$
  • C. $$3$$
  • D. $$\frac{1}{4}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Medium
Let $$n \space\epsilon \space N$$ & the A.M., G.M., H.M. & the root mean square of $$n$$ numbers $$2n+1, 2n+2, ...,$$ up to $$n^{th}$$ number are $$A_{n}$$, $$G_{n}$$, $$H_{n}$$ and $$R_{n}$$ respectively. 
On the basis of above information answer the following questions

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer