Mathematics

# Evaluate the following integral:$\int { \cfrac { \sin ^{ 3 }{ x } }{ \sqrt { \cos { x } } } } dx\quad$

##### SOLUTION
$\displaystyle\int{\dfrac{{\sin}^{3}{x}}{\sqrt{\cos{x}}}dx}$

$=\displaystyle\int{\dfrac{\left(1-{\cos}^{2}{x}\right)\sin{x}}{\sqrt{\cos{x}}}dx}$

Let $t=\cos{x}\Rightarrow\,dt=-\sin{x}\,dx$

$=-\displaystyle\int{\dfrac{\left(1-{t}^{2}\right)dt}{\sqrt{t}}}$

$=-\displaystyle\int{\dfrac{dt}{\sqrt{t}}}+\displaystyle\int{\dfrac{{t}^{2}\,dt}{\sqrt{t}}}$

$=-\displaystyle\int{{t}^{-\frac{1}{2}}dt}+\displaystyle\int{{t}^{2-\frac{1}{2}}\,dt}$

$=-\displaystyle\int{{t}^{-\frac{1}{2}}dt}+\displaystyle\int{{t}^{\frac{3}{2}}\,dt}$

$=-\dfrac{{t}^{-\frac{1}{2}+1}}{-\dfrac{1}{2}+1}+\dfrac{{t}^{\frac{3}{2}+1}}{\dfrac{3}{2}+1}+c$

$=-\dfrac{{t}^{\frac{1}{2}}}{\dfrac{1}{2}}+\dfrac{{t}^{\frac{5}{2}}}{\dfrac{5}{2}}+c$

$=-{2}{t}^{\frac{1}{2}}+\dfrac{2}{5}{t}^{\frac{5}{2}}+c$

$=-{2}{\cos}^{\frac{1}{2}}{x}+\dfrac{2}{5}{cos}^{\frac{5}{2}}{x}+c$ where

$t=\cos{x}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
Evaluate the definite integral, $\int\limits_{ - 1}^1 {\dfrac{{\left( {2{x^{332}} + {x^{998}} + 4{x^{1668}} \cdot \sin {x^{691}}} \right)}}{{1 + {x^{666}}}}dx}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int { \dfrac { dx }{ { x }^{ 2 }\sqrt { { x }^{ 2 }+{ a }^{ 2 } } } = }$

• A. $\dfrac { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } }{ { a }^{ 2 }x } +c$
• B. $-\dfrac { { a }^{ 2 }x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } +c$
• C. $\dfrac { { a }^{ 2 }x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } +c$
• D. $-\dfrac { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } }{ { a }^{ 2 }x } +c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate: $\displaystyle \int \frac{\log \left [ x+\sqrt{1+x^{2}} \right ]}{\sqrt{1+x^{2}}}dx$
• A. $\displaystyle \left [ \log \left ( x+\sqrt{1+x^{2}} \right ) \right ]^{2}$
• B. $\displaystyle 2\left [ \log \left ( x+\sqrt{1+x^{2}} \right ) \right ]^{2}$
• C. $\displaystyle \frac{1}{2}\left [ \log \left ( x+\sqrt{1+x^{2}} \right ) \right ]$
• D. $\displaystyle \frac{1}{2}\left [ \log \left ( x+\sqrt{1+x^{2}} \right ) \right ]^{2}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate: $\displaystyle \int {\dfrac {dx } { \sqrt { 1 - {e} ^ { 2 x } } } }$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts
$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
where  $\displaystyle v_{1}\left ( x \right )=\int v\left ( x \right )dx,v_{2}\left ( x \right )=\int v_{1}\left ( x \right )dx ..., v_{n}\left ( x \right )= \int v_{n-1}\left ( x \right )dx$
Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020