Mathematics

# $\displaystyle \int \dfrac {1}{\sqrt {\sin^{3}x\sin(x+a)}}dx$ is equal to

$-2\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \cot x}+c$

##### SOLUTION
$\begin{array}{l} \int { \frac { { dx } }{ { \sqrt { { { \sin }^{ 3 } }x\sin \left( { x+a } \right) } } } } \\ =\int { \frac { { dx } }{ { \sqrt { { { \sin }^{ 3 } }x\sin x\left( { \cos a+\cot x\sin a } \right) } } } } \\ =\int { \frac { { \cos e{ c^{ 2 } }xdx } }{ { \sqrt { \cos a+\sin a\cot x } } } } \\ \cos a+\sin a\cot =t \\ \frac { { dt } }{ { dx } } =-\sin a\cos e{ c^{ 2 } }x \\ -\frac { 1 }{ { \sin a } } dt=\cos e{ c^{ 2 } }xdx \\ =-\frac { 1 }{ { \sin a } } \int { \frac { { dt } }{ { \sqrt { t } } } } \\ =-\frac { 1 }{ { \sin a } } \int { { t^{ \frac { { -1 } }{ 2 } } }dt } \\ =\frac { { -2 } }{ { \sin a } } { t^{ \frac { 1 }{ 2 } } }+C \\ =-2\cos eca\sqrt { \cos a+\sin a\cot x } +C \end{array}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
In $I_n=\displaystyle\int (ln x)^ndx$, then $I_n+nI_{n-1}=?$
• A. $\dfrac{(ln x)^n}{x}+C$
• B. $x(ln x)^{n-1}+C$
• C. None of these
• D. $x(ln x)^n+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate the given integral.
$\displaystyle \int { \left( \cfrac { 2-x }{ { (1-x) }^{ 2 } } \right) } { e }^{ x }dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Solve: $\displaystyle \int\dfrac{x + 2}{2x^2 + 6x + 5}dx$
• A. $\frac{1}{2}log(2x^2+6x+5)+\frac{1}{2}tan^{-1}(2x+3)+C$
• B. $\frac{1}{4}log(2x^2+6x+5)-\frac{1}{2}tan^{-1}(2x+3)+C$
• C. $\frac{1}{2}log(2x^2+6x+5)-\frac{1}{2}tan^{-1}(2x+3)+C$
• D. $\frac{1}{4}log(2x^2+6x+5)+\frac{1}{2}tan^{-1}(2x+3)+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\displaystyle \int \dfrac{x^2}{2+x^3} dx$

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