Mathematics

# $\int e^{2ax}\dfrac {1-\cos 2a x}{1+\sin 2ax}dx$ is equal to

$-\dfrac {1}{a}e^{2ax}\cot \left(\dfrac {\pi}{4}+ax\right)+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
Solve $\displaystyle\int {\dfrac{{dx}}{{\sqrt {x - {x^2}} }}}$
• A. $I={{\sin }^{-1}}\left( 2x+1 \right)+C$
• B. $I={{\sin }^{-1}}\left( x-1 \right)+C$
• C. None of these
• D. $I={{\sin }^{-1}}\left( 2x-1 \right)+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Integrate:$\displaystyle \int _{ 0 }^{ \pi /2 }{ \cos ^{ 2 }{ x } dx }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Solve $\displaystyle\int \dfrac {\tan x}{(a^{2} + b^{2}\tan^{2}x)} dx$.
• A. $I= \dfrac{1}{{5\left( {{b^2} - {a^2}} \right)}}\log \left| {{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x} \right| + C$
• B. $I= \dfrac{1}{{2\left( {{b^2} - {a^2}} \right)}}\log \left| {{a^2}{{\cos }^2}x - {b^2}{{\sin }^2}x} \right| + C$
• C. None of these
• D. $I= \dfrac{1}{{2\left( {{b^2} - {a^2}} \right)}}\log \left| {{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x} \right| + C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate : $\displaystyle \int \dfrac{\sin^{8} x- \cos^{8} x}{1-2 \sin^{2} x \cos^{2} x} dx$

Solve $\displaystyle\int\dfrac{{{{\left( {\log \,x} \right)}^2}}}{x}$