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 $\sin (A-B)=\sin A\cos B-\cos A\sin B$]
$I=\displaystyle\int \dfrac{[\sin (x+\alpha)\cos (2\alpha)-\cos(x\alpha)\sin (2\alpha)]}{\sin (x+\alpha)}dx$
$I=\displaystyle\int \cos 2\alpha dx-intn \cot (x+\alpha).\sin 2\alpha dx$
$I=\cos 2\alpha.x-\ln |\sin (x+\alpha)|.\sin 2\alpha$
$I=x.\cos 2\alpha-\sin 2\alpha.\ln |\sin (x+\alpha)|+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### 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. $I_{1}=I_{2}$

1 Verified Answer | Published on 17th 09, 2020

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

1 Verified Answer | Published on 17th 09, 2020

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

1 Verified Answer | Published on 17th 09, 2020

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. $\displaystyle \pi /2$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$