Mathematics

$$\displaystyle\int \dfrac{\sin (x-\alpha)}{\sin (x+\alpha)}dx$$.


SOLUTION
We are given, $$I=\displaystyle\int\dfrac{\sin (x-\alpha)}{\sin (x+a)}dx$$
$$I=\displaystyle\int \dfrac{\sin (x+\alpha-2\alpha)}{\sin (x+\alpha)}dx$$
[ Now we know that $$\sin (A-B)=\sin A\cos B-\cos A\sin B$$]
$$I=\displaystyle\int \dfrac{[\sin (x+\alpha)\cos (2\alpha)-\cos(x\alpha)\sin (2\alpha)]}{\sin (x+\alpha)}dx$$
$$I=\displaystyle\int \cos 2\alpha dx-intn \cot (x+\alpha).\sin 2\alpha dx$$
$$I=\cos 2\alpha.x-\ln |\sin (x+\alpha)|.\sin 2\alpha$$
$$I=x.\cos 2\alpha-\sin 2\alpha.\ln |\sin (x+\alpha)|+C$$
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 Single Correct Medium
If $$I_{1}=\displaystyle \int_{e}^{e^{2}}\dfrac{dx}{\ell n x}$$ and $$I_{2}=\int_{1}^{2} \dfrac{e^{x}}{x}dx$$, then 
  • A. $$2I_{1}=I_{2}$$
  • B. $$I_{1}=2I_{2}$$
  • C. $$I_{1}+I_{2}=0$$
  • D. $$I_{1}=I_{2}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Prove that $$\displaystyle \int_{0}^{\pi /4}\frac{dx}{\cos ^{4}x-\cos ^{2}x\sin ^{2}x+\sin ^{4}x}=\frac{\pi }{2}$$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
$$\int _{ -5 }^{ 5 }{ \left(x+2\right)} dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Hard
$$\displaystyle \int_{-\pi /2}^{\pi /2}\frac{\pi ^{\sin x}}{1+\pi ^{\sin x}}dx$$
  • A. $$0$$
  • B. $$\displaystyle \pi /4$$
  • C. $$\displaystyle \pi $$
  • D. $$\displaystyle \pi /2$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Easy
Evaluate:
$$ \int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer