Mathematics

# $\int { \cfrac { \csc ^{ 2 }{ x } -2005 }{ \cos ^{ 2005 }{ x } } dx }$ is equal to

$\cfrac { \cot { x } }{ { \left( \cos { x } \right) }^{ 2005 } } +C$

##### SOLUTION
$\displaystyle\int \dfrac{\text{cosec}^{2}x-2005}{\cos^{2005}x}dx$
Put $t=\dfrac{\cot x}{\cos ^{2005}x}\implies dt=\dfrac{\text{cosec}^{2}x(\cos^{2005}x)-(\cot x)(2005\cos^{2004}x\sin x)}{(\cos^{2005}x)^{2}}dx$
$=\dfrac{\text{cosec}^{2}x-2005}{\cos^{2005}x}dx$
$\displaystyle\int dt=t+C=\dfrac{\cot x}{\cos^{2005}x}+C$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int x \tan^{-l} (x^{2})dx=$
• A. $\displaystyle \frac{1}{2}[x^{2}\tan^{-1}(x^{2})+\log(1+x^{4})]+c$
• B. $x^{2}\tan^{-1}(x^{2})-\displaystyle \frac{1}{2}\log(1+x^{4})+c$
• C. $x^{2}\tan^{-1}(x^{2})+\displaystyle \frac{1}{2}\log(1+x^{4})+c$
• D. $\displaystyle \frac{1}{2}[x^{2}\tan^{-1}(x^{2})-\frac{1}{2}\log(1+x^{4})]+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$I = \displaystyle \int \sec x\tan x dx$ is
equal to
• A. $\cos x+c$
• B. $\tan x+c$
• C. None of these
• D. $\sec x+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\displaystyle\int \dfrac{dx}{\sqrt{10-10x-2x^2}}$=?

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int\frac{dx}{x(1+\sqrt[3]{x})^{2}}$ is equal to
• A. $3(\displaystyle \log\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}+\frac{1}{1+\sqrt[3]{x}})+c$
• B. $3(\displaystyle \log\frac{x^{1/3}}{1+x^{1/3}}-\frac{1}{1+\sqrt[3]{x}})+c$
• C. $3(\displaystyle \log\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}-\frac{1}{1+\sqrt[3]{x}})+c$
• D. $3(\displaystyle \log\frac{x^{1/3}}{1+x^{1/3}}+\frac{1}{1+\sqrt[3]{x}})+c$

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