Mathematics

# Solve : $\dfrac{dx}{\sin^3x .\cos^5x}$

##### SOLUTION
$I=\int { \dfrac { dx }{ \left( { \ sin }^{ 3 }x.{ \ cos }^{ 5 }x \right) } } \\$ substituting,\\ $\sin x=\dfrac { \tan x }{ \sec x } \\ \cos x\dfrac { 1 }{ \sec x } \\ { \sec }^{ 2 }x={ \tan }^{ 2 }x+1\\$ $\therefore I=\int { { \sec }^{ 2 }x } .\dfrac { \left( \tan^{ 2 }x+1 \right) ^{ 3 } }{ { \tan }^{ 3 }x } dx\\ Let,\\ \quad \quad u=\tan x\\$ $\rightarrow dx=\dfrac { 1 }{ { \sec }^{ 2 } } du\\$ $\therefore I=\int { \dfrac { { \left( { u }^{ 2 }+1 \right) }^{ 3 } }{ { u }^{ 3 } } du } \\ \quad \quad =\int { \left( { u }^{ 3 }+3u+\dfrac { 3 }{ u } +\dfrac { 1 }{ { u }^{ 3 } } \right) du }$ \\ $\quad \quad =\int { { u }^{ 3 }du } +3\int { udu } +3\int { \dfrac { 1 }{ u } du } +\int { \dfrac { 1 }{ { u }^{ 3 } } du }$ \\ $\quad \quad =3\log u+\dfrac { { u }^{ 4 } }{ 4 } +\dfrac { { 3u }^{ 2 } }{ 2 } -\dfrac { 1 }{ { 2u }^{ 2 } }$
=$3 \log \ tanx$$+ \dfrac { { \tan }^{ 4 }x }{ 4 } +\dfrac { 3{ \tan }^{ 2 }x }{ 2 } -\dfrac { 1 }{ 2{ tan }^{ 2 }x } +C$

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

#### Realted Questions

Q1 Single Correct Medium

$\displaystyle \frac{x^{2}+2x+3}{x^{3}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}\Rightarrow A+B-C=$
• A. 6
• B. 3
• C. 1
• D.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int e^{x}(\tanh x+\sec h^{2}x)dx=$
• A. $e^{x}sec h^{2} x+c$ \$
• B. $e^{x}sinh\ x+c$
• C. $e^{x}$cosec$h^{2}x+c$
• D. $e^{x}\tan hx+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\int \dfrac {1}{1 + \sin x}dx = \tan \left (\dfrac {x}{2} + a\right ) + b$, then
• A. $a = -\dfrac {\pi}{4}, b\in R$
• B. $a = \dfrac {5\pi}{4}, b\in R$
• C. None of these
• D. $a = \dfrac {\pi}{4}, b\in R$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle \int \sqrt{e^{x}-1}\:dx$ is equal to
• A. $\displaystyle \sqrt{e^{x}-1}-\tan^{-1}\sqrt{e^{x}-1}+C$
• B. $\displaystyle \sqrt{e^{x}-1}+\tan^{-1}\sqrt{e^{x}-1}+C$
• C. $\displaystyle 2\left [ \sqrt{e^{x}-1}+\tan^{-1}\sqrt{e^{x}-1} \right ]+C$
• D. $\displaystyle 2\left [ \sqrt{e^{x}-1}-\tan^{-1}\sqrt{e^{x}-1} \right ]+C$

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 $\displaystyle f(x)>0 \forall x \in R$