Mathematics

# $\displaystyle \int \dfrac{1}{\sqrt{x}} \tan^4 \, \sqrt{x} \, \sec^2 \, \sqrt{x} \, dx =$

$\dfrac{2}{5} \tan^5 \, \sqrt{x} + c$

##### SOLUTION
$\displaystyle \int \dfrac{\tan ^4\sqrt{x}\sec ^2\sqrt{x}}{\sqrt{x}}dx$

substitute $u=\tan \sqrt{x}\rightarrow du=\dfrac{\sec ^2\sqrt{x}}{2\sqrt{x}}dx$

$=2\int u^4du$

$=\dfrac{2}{5}u^5$

$=\dfrac{2}{5}\tan ^5 \sqrt{x}+C$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\int \frac{x}{1+\cos x}dx.$
• A. $\displaystyle \tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).$
• B. $\displaystyle x\tan \left ( x/2 \right )+2\log \sec \left ( x/2 \right ).$
• C. $\displaystyle x\tan \left ( x/2 \right )-\log \sec \left ( x/2 \right ).$
• D. $\displaystyle x\tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate:
$\displaystyle\int \dfrac{x^8+8}{x^4-2x+4}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate:$\int \frac{1}{x^{4}-1}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve: $\int \sin^32x.dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
The value of $\displaystyle \int_0^{\dfrac{\pi}{2}}\dfrac{\sqrt{\cot t}}{\sqrt{\cot t}+\sqrt{\tan t}}dt$
• A. $\dfrac{\pi}{2}$
• B. $\dfrac{\pi}{6}$
• C. $\dfrac{\pi}{8}$
• D. $\dfrac{\pi}{4}$