Mathematics

Solve: $$\displaystyle \int \dfrac{\cos x}{(a+b\sin x)^2}dx$$


SOLUTION
Now,
 $$\displaystyle \int \dfrac{\cos x}{(a+b\sin x)^2}dx$$
 $$=\dfrac{1}{b}\displaystyle \int \dfrac{b\cos x}{(a+b\sin x)^2}dx$$
 $$=\dfrac{1}{b}\displaystyle \int \dfrac{d(a+b\sin x)}{(a+b\sin x)^2}$$
$$=-\dfrac{1}{b}.\dfrac{1}{a+b\sin x }+c$$ [ Where $$c$$ is integrating constant]
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 84
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
Solve $$\displaystyle\int \dfrac{x^3+4x^2-7x+5}{x+2}dx$$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium
Value of the definite integral $$\displaystyle \int_{-1}^{1}\dfrac {dx}{(1+x^{3}+\sqrt {1+x^{6}})}$$
  • A. $$is\ 1$$
  • B. $$is\ 2$$
  • C. $$is\ zero$$
  • D. $$is\dfrac {1}{2}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Hard
The integral $$\displaystyle \int { \left( 1+2{ x }^{ 2 }+\frac { 1 }{ x }  \right)  } { e }^{ { x }^{ 2-\frac { 1 }{ x }  } }dx$$ is equal to
  • A. $$(2x-1).e^{x^{2-\dfrac {1}{x}}}+c$$
  • B. $$(2x+1).e^{x^{2-\dfrac {1}{x}}}+c$$
  • C. $$-xe^{x^{2-\dfrac {1}{x}}}+c$$
  • D. $$xe^{x^{2-\dfrac {1}{x}}}+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Solve $$\int {\frac{1}{\cos ^2x(1-\tan x)^2}dx}$$

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