Mathematics

# Prove that $\displaystyle\int^{\tan x}_{1/e}\dfrac{t}{1+t^2}dt+\displaystyle\int^{\cot x}_{1/e}\dfrac{1}{t(1+t^2)}dt=1$.

##### SOLUTION
To prove $\int _{ \cfrac { 1 }{ e } }^{ \tan { x } }{ \cfrac { t }{ 1+{ t }^{ 2 } } dt } +\int _{ \cfrac { 1 }{ e } }^{ \cot { x } }{ \cfrac { 1 }{ t(1+{ t }^{ 2 }) } dt }$
Let ${ I }_{ 1 }=\int _{ \cfrac { 1 }{ e } }^{ \tan { x } }{ \cfrac { t }{ 1+{ t }^{ 2 } } dt } \\ { I }_{ 2 }=\int _{ \cfrac { 1 }{ e } }^{ \cot { x } }{ \cfrac { 1 }{ t(1+{ t }^{ 2 }) } dt }$
and $I={ I }_{ 1 }+{ I }_{ 2 }$
also, $\int _{ \cfrac { 1 }{ e } }^{ \cot { x } }{ \cfrac { 1 }{ t(1+{ t }^{ 2 }) } dt }$
Let $t=\cfrac { 1 }{ z }$
$dt=\cfrac { -1 }{ { z }^{ 2 } } dz$
So, ${ I }_{ 1 }=\int _{ e }^{ \tan { x } }{ \cfrac { -zdz }{ { z }^{ 2 }(1+\cfrac { 1 }{ { z }^{ 2 } } ) } } =\int _{ \tan { x } }^{ e }{ \cfrac { zdz }{ 1+{ z }^{ 2 } } }$
So, ${ I }_{ 2 }=\int _{ \tan { x } }^{ e }{ \cfrac { tdt }{ 1+{ t }^{ 2 } } }$
So, $I={ I }_{ 1 }+{ I }_{ 2 }$
$I=\int _{ \cfrac { 1 }{ e } }^{ e }{ \cfrac { tdt }{ 1+{ t }^{ 2 } } } \\ =\cfrac { 1 }{ 2 } \int _{ \cfrac { 1 }{ e } }^{ e }{ \cfrac { 2t }{ 1+{ t }^{ 2 } } dt } \\ I=\cfrac { 1 }{ 2 } { \log { (1+{ t }^{ 2 }) } }_{ \cfrac { 1 }{ e } }^{ e }\\ I=\cfrac { 1 }{ 2 } \log { (1+{ e }^{ 2 }) } -\cfrac { 1 }{ 2 } \log { (1+\cfrac { 1 }{ { e }^{ 2 } } ) } \\ I=\cfrac { 1 }{ 2 } [\log { (\cfrac { 1+{ e }^{ 2 } }{ 1+\cfrac { 1 }{ { e }^{ 2 } } } ) } ]\\ I=\cfrac { 1 }{ 2 } \log { { e }^{ 2 } } \\ I=\cfrac { 1 }{ 2 } \times 2\\ I=1$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle\int^{10}_0f(x)dx=5$, then $\displaystyle\sum^{10}_{k=1}\displaystyle\int^1_0f(k-1+x)dx$ is?
• A. $50$
• B. $10$
• C. None of these
• D. $5$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\displaystyle\int_{0}^{2} (1+2x)dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of integral $\displaystyle \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \cfrac { \sin { x } -x\cos { x } }{ x(x+\sin { x } ) } } dx$ is
• A. $\log _{ e }{ \left\{ \cfrac { (\pi +3) }{ 2\left( 2\pi +3\sqrt { 3 } \right) } \right\} }$
• B. $\log _{ e }{ \left\{ \cfrac { \left( 2\pi +3\sqrt { 3 } \right) }{ 2(\pi +3) } \right\} }$
• C. $\log _{ e }{ \left\{ \cfrac { 2\left( 2\pi +3\sqrt { 3 } \right) }{ (\pi +3) } \right\} }$
• D. $\log _{ e }{ \left\{ \cfrac { 2(\pi +3) }{ \left( 2\pi +3\sqrt { 3 } \right) } \right\} }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
lf $\displaystyle \int\frac{dx}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}}=A\sqrt [4]{\frac{x-1}{x+2}}+c$, then $A$ is equal to
• A. $\dfrac{1}{3}$
• B. $\dfrac{2}{3}$
• C. $\dfrac{3}{4}$
• D. $\dfrac{4}{3}$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$