Mathematics

Evaluate the following integral:
$$\int { \cfrac { \sin { 2x }  }{ { \left( a+b\cos { 2x }  \right)  }^{ 2 } }  } dx$$


SOLUTION
Let 
$$t=a+b\cos{2x}\Rightarrow\,dt=-2b\sin{2x}dx$$

$$\Rightarrow\,\dfrac{-1}{2b}dt=\sin{2x}dx$$

Now,$$\displaystyle\int{\dfrac{\sin{2x}}{{\left(a+b\cos{2x}\right)}^{2}}dx}$$

$$=\dfrac{-1}{2b}\displaystyle\int{\dfrac{dt}{{t}^{2}}}$$

$$=\dfrac{-1}{2b}\displaystyle\int{{t}^{-2}dt}$$

$$=\dfrac{-1}{2b}\left[\dfrac{{t}^{-2+1}}{-2+1}\right]+c$$

$$=\dfrac{1}{2bt}+c$$

$$=\dfrac{1}{2b\left(a+b\cos{2x}\right)}+c$$ where $$t=a+b\cos{2x}$$
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 86
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
If $$\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { \sin { x }  }{ 1+\sin { x+\cos { x }  }  }  }dx=\dfrac{\pi}{a}-\dfrac{1}{2}lnb,\ a,b \in N$$ then
  • A. $$a+b=4$$
  • B. $$a-b=4$$
  • C. $$a-b=6$$
  • D. $$a+b=6$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 One Word Hard
Evaluate $$\displaystyle \int \frac{e^{x}}{e^{2x}+6e^{x}+5}dx.$$
The ans is $$\displaystyle =\frac{1}{4}log\left | \frac{e^{x}+1}{e^{x}+ k} \right |+C$$
what is k

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Hard
If $$\displaystyle \int _{ \frac { 1 }{ \sqrt { 3 }  }  }^{  \frac { 1 }{ \sqrt { 3 }  }  }{\displaystyle  \frac { x^{ 4 } }{ 1-x^{ 4 } } \ \cos ^{ -1 }{\displaystyle  \frac { 2x }{ 1+x^{ 2 } }  } dx } =k\int _{ 0 }^{  \frac { 1 }{ \sqrt { 3 }  }  }{ \displaystyle \frac { x^{ 4 } }{ 1-x^{ 4 } } dx } $$, 
then $$'k'$$ is equal to?
  • A. $$2\pi$$
  • B. $$2$$
  • C. $$1$$
  • D. $$\pi$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Hard
If $$\displaystyle I = \int \frac {\sec^{-1} \sqrt x - cosec^{-1} \sqrt x}{\sec^{-1} \sqrt x + cosec^{-1} \sqrt x} dx$$, then I equals
  • A. $$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) + x + C$$
  • B. $$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x + \sqrt {x - 1}) - x + C$$
  • C. none of these
  • D. $$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) - x + C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Hard
Evaluate: $$\displaystyle\int \tan^{-1}\sqrt{x}dx$$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer