Mathematics

# Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number $n$.If $\displaystyle I_n = \int e^{\alpha \: x} \sin^n x \: dx = \frac { e^{\alpha \: x}}{\alpha^2 + n} \sin^{n - 1} x (\alpha \: \sin \: x - n \: \cos \: x) + A \int e^{\alpha x}\sin^{n-2}x\cos x$, then A is equal to

$\displaystyle \frac {n(n - 1)e^{\alpha x}}{\alpha^2 + n^2}$

##### SOLUTION
$I_{n}=\displaystyle\int e^{\alpha x}\sin^{n} x d{x}=\dfrac{e^{\alpha x}}{\alpha}\sin^{n} x-\dfrac{n}{\alpha}\int e^{\alpha x}\sin^{n-1}x\cos x d {x}$
$I_{n}=\dfrac{e^{\alpha x}}{\alpha}\sin^{n}x-\dfrac{n}{\alpha^2}e^{\alpha x}\sin^{n-1}x\cos x+\dfrac{n}{\alpha ^2}\displaystyle\int [(n-1)\sin ^{n-2}x\cos^2 x-\sin^{n}x]e^{\alpha x} d{x}$
$I_{n}\bigg(1+\dfrac{n^2}{\alpha^2}\bigg)=\dfrac{e^{\alpha x}}{\alpha^2}\sin^{n-1}x [\alpha \sin x-n\cos x]+\dfrac{n(n-1)}{\alpha^2}\displaystyle\int e^{\alpha x}\sin^{n-2}x d{x}$
$I_{n}=\dfrac{e^{\alpha x}}{\alpha^2+n^2}\sin^{n-1}x[\alpha \sin x-n\cos x]+\dfrac{n(n-1)}{\alpha^2+n^2}I_{n-2}$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Hard
The value of $\displaystyle \int { \sqrt { \frac { a-x }{ a+x } dx } }$ is
• A. $\displaystyle a{ \sin }^{ -1 }\left( \frac { x }{ a } \right) -\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +c$
• B. $\displaystyle a{ \cos }^{ -1 }\left( \frac { x }{ a } \right) +\sqrt { { a }^{ 2 }+{ x }^{ 2 } } +c$
• C. $\displaystyle { \sin }^{ -1 }\left( \frac { x }{ a } \right) +\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +c$
• D. $\displaystyle a{ \sin }^{ -1 }\left( \frac { x }{ a } \right) +\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate: $\displaystyle\int_{1}^{2} \dfrac 2x\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int {\frac{{{{\sin }^{ - 1}}x}}{{{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}}}dx}$
• A. $\frac{1}{2}\log \left| {\left( {1 - {x^2}} \right)} \right| + C$
• B. $\frac{{x\left( {{{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }}+C$
• C. $4+\frac { \pi }{ 2 }$
• D. $\frac{{x\left( {{{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }} + \frac{1}{2}\log \left| {\left( {1 - {x^2}} \right)} \right| + C.$

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Hard
The value of the integral $\displaystyle \int ^{\pi /4}_{-\pi /4}\displaystyle \frac{dx}{a^{2}\cos ^{2}x+b^{2}\sin ^{2}x}$
• A. $\displaystyle \frac{2}{ab}\tan ^{-1}\displaystyle \frac{a}{b}+\displaystyle \frac{1}{ab}$
• B. $\displaystyle \frac{2}{ab}\tan ^{-1}\displaystyle \frac{b}{a}\left ( a> 0, b> 0 \right )$
• C. $\displaystyle \frac{2}{ab}\tan ^{-1}\displaystyle \frac{b}{a}\left ( a< 0, b< 0 \right )$
• D. $\displaystyle \frac{\pi }{2}\left ( a= 1, b= 1 \right )$

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