Mathematics

# Write a value of $\int { \tan ^{ 6 }{ x } } \sec ^{ 2 }{ x } dx$

##### SOLUTION
Let $t=\tan{x}\Rightarrow\,dt={\sec}^{2}{x}dx$

$I=\displaystyle\int{{\tan}^{6}{x}{\sec}^{2}{x}dx}$

$=\displaystyle\int{{t}^{6}dt}$

$=\dfrac{{t}^{7}}{7}+c$

$=\dfrac{{\tan}^{7}{x}}{7}+c$    ........where $t=\tan{x}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
Evaluate  :    $I = \int {\dfrac{{x + 9}}{{{x^2} + 5}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate : $\displaystyle \int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\times \left [ \sin ^{-1}\left ( 1-2x^{2} \right ) +\cos ^{-1}\left ( 2x\sqrt{1-x^{2}} \right )\right ]dx$
• A. $\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$
• B. $\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$
• C. $\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$
• D. $\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate
$\int _{ -1 }^{ 1 }{ { x }^{ 17 } } { cos }^{ 4 }xdx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\displaystyle\int f(x)dx=g(x),$ then $\displaystyle\int f^{-1}(x)dx=$ _____________$+$c.
• A. $xf^{-1}(x)-g(f^{-1}(x))$
• B. $xf^{-1}(x)-g(f(x))$
• C. $x\cdot f^{-1}(x)$
• D. $x\cdot f(x)-g(f^{-1}(x))$

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