Mathematics

# The value of $\displaystyle \int_{0}^{\pi /4}\sqrt{\tan \theta }\: d\theta$ is?

None of the above

##### SOLUTION
Let $\displaystyle I=\int { \sqrt { \tan { \theta } } d\theta } =\frac { 1 }{ 2 } \left( { I }_{ 1 }-{ I }_{ 2 } \right)$

Where
$\displaystyle { I }_{ 1 }=\int { \left( \tan { \theta } +\cot { \theta } \right) d\theta } =\int { \frac { \sin { \theta } +\cos { \theta } }{ \sqrt { \cos { \theta } \sin { \theta } } } d\theta }$
Put $\sin { \theta } +\cos { \theta } =u$
$\displaystyle { I }_{ 1 }=\sqrt { 2 } \int { \frac { dt }{ \sqrt { { 1-t }^{ 2 } } } } =\sqrt { 2 } \sin ^{ -1 }{ t } \\ =\sqrt { 2 } \sin ^{ -1 }{ \left( \sin { \theta } -\cos { \theta } \right) }$

And  ${ I }_{ 2 }=\int \left( \sqrt { \cot \theta } -\sqrt { \tan \theta } \right) d\theta$

$\displaystyle =\int { \frac { \cos { \theta } -\sin { \theta } }{ \sqrt { \cos { \theta } \sin { \theta } } } d\theta }$

Put $\sin { \theta } +\cos { \theta } =t\Rightarrow 2\sin { \theta } \cos { \theta } ={ t }^{ 2 }-1$
$\displaystyle { I }_{ 2 }=\sqrt { 2 } \int { \frac { dt }{ \sqrt { { t }^{ 2 }-1 } } } =\sqrt { 2 } \log { \left( 1+\sqrt { { t }^{ 2 }-1 } \right) } \\ =\sqrt { 2 } \log { \left( \sin { \theta } +\cos { \theta } +\sqrt { \sin { 2\theta } } \right) } \\ I=\sqrt { 2 } \sin ^{ -1 }{ \left( \sin { \theta } -\cos { \theta } \right) } +\sqrt { 2 } \log { \left( \sin { \theta } +\cos { \theta } +\sqrt { \sin { 2\theta } } \right) }$

Therefore
$\displaystyle \int _{ 0 }^{ \pi /4 }{ Id\theta } =\sqrt { 2 } \log { \left( \sqrt { 2 } +1 \right) } -\frac { 3\pi }{ 2 }$

Its FREE, you're just one step away

Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 One Word Medium
Prove that $\displaystyle \int\frac{e^{\log \left ( 1+1/x^{2} \right )}}{x^{2}+1/x^{2}}dx=\frac{1}{\sqrt{\left ( 2 \right )}}\tan ^{-1}\left ( x-\frac{1}{x} \right )$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Resolve $\displaystyle \frac{1}{x^4+1}$ into partial fractions.

• A. $\displaystyle \frac{(x+\sqrt{2})}{4\sqrt{2}(x^2+x\sqrt{2}+1)}-\frac{(x-\sqrt{2})}{4\sqrt{2}(x^2-x\sqrt{2}+1)}$
• B. $\displaystyle \frac{(x+\sqrt{2})}{\sqrt{2}(x^2+x\sqrt{2}+1)}-\frac{(x-\sqrt{2})}{\sqrt{2}(x^2-x\sqrt{2}+1)}$
• C. $\displaystyle -\frac{(x+\sqrt{2})}{2\sqrt{2}(x^2+x\sqrt{2}+1)}+\frac{(x-\sqrt{2})}{2\sqrt{2}(x^2-x\sqrt{2}+1)}$
• D. $\displaystyle \frac{(x+\sqrt{2})}{2\sqrt{2}(x^2+x\sqrt{2}+1)}-\frac{(x-\sqrt{2})}{2\sqrt{2}(x^2-x\sqrt{2}+1)}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Multiple Correct Hard
Let $\displaystyle \alpha +\beta =1,2\alpha ^{2}+2\beta ^{2}=1$ and f(x) be a continuous function such that f(x + 2) + f(x) = 2 $\displaystyle \forall \times \epsilon \left [ 0,2 \right ]\: \: and\: \: \left ( p+4 \right )=\int_{0}^{4}f\left ( x \right )dx\: \&\: q=\frac{\alpha }{\beta }$ exactly one root of the equation $\displaystyle ax^{2}-bx+c=0$ is lying between p and q when a, b, $\displaystyle c\: \epsilon \: N$ then
• A. $\displaystyle b^{2}-4ac\leq 0$
• B.
$c(a - b + c) > 0$
• C. $\displaystyle b^{2}-4ac\geq 0$
• D. $c(a - b + c) < 0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate $\displaystyle \int \dfrac{1}{\sqrt{25+9x^{2}}}dx$

$\int \frac{1}{1+x}\;dx$