Mathematics

# $\int e^{\tan \theta}(\sec \theta -\sin \theta)d\theta$ equals

$e^{\tan \theta}\cos \theta+c$

##### SOLUTION
Let $\displaystyle I=\int { { e }^{ \tan { \theta } } } \left( \sec { \theta } -\sec { \theta } \right) d\theta$
Substitute $\tan { \theta } =t\Rightarrow \sec^{ 2 }{ \theta }d\theta =dt$
$\therefore I=\int { { e }^{ t } } \left( \cos { \theta } -\tan { \theta } \sec { \theta } \right) dt$
$\displaystyle =\int { { e }^{ t } } \left( \frac { 1 }{ \sqrt { 1+{ t }^{ 2 } } } -t\sqrt { 1+{ t }^{ 2 } } \right) dt$
As $\displaystyle \frac { d }{ dx } \left( \frac { 1 }{ \sqrt { 1+{ t }^{ 2 } } } \right) =t\sqrt { 1+{ t }^{ 2 } }$
$\displaystyle \therefore I={ e }^{ t }\left( \frac { 1 }{ \sqrt { 1+{ t }^{ 2 } } } \right) ={ e }^{ \tan { \theta } }\cos { \theta } +c$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
Prove that $\displaystyle\int^{\pi/2}_0\dfrac{\sin^3x}{(\sin^3x+\cos^3x)}dx=\dfrac{\pi}{4}$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve :
$\displaystyle \int_{0}^{\dfrac {\pi}{2}} \sin^3 xdx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The antiderivative of every odd function is
• A. an odd function
• B. neither even nor odd
• C. sometimes even, sometimes odd
• D. an even function

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $A=\displaystyle \int_{0}^{1}{\dfrac{{e}^{t}}{1+t}}dt$ then $\displaystyle \int_{0}^{1}{{e}^{t}ln(1+t)}dt=$
• A. $e\ ln{2}+A$
• B. $Ae\ ln{2}$
• C. $A\ ln{2}$
• D. $e\ ln{2}-A$

$\underset {n\rightarrow \infty}{lim}\dfrac{1^2+2^2+3^2+.....+n^2}{n^3}=.................$