Mathematics

# Find Integrals of given function: $\int_{}^{} {\tan \theta } {\tan ^2}\theta {\sec ^2}\theta d\theta$

$\dfrac{\tan^4\theta}{4}+c$

##### SOLUTION
$\int_{}^{} {\tan \theta } {\tan ^2}\theta {\sec ^2}\theta d\theta=\int\tan^3\theta\sec^2\theta d\theta$

Let $t=\tan\theta\\dt=\sec^2\theta d\theta$

Hence,
$\int t^3dt=\dfrac{t^4}{4}+c=\dfrac{\tan^4\theta}{4}+c$

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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
If $\displaystyle \frac{x^{3}}{(x-a)(x-b)(x-c)}=1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$ then $\mathrm{A}=$
• A. $\displaystyle \frac{a^{3}}{(c-b)(c-a)}$
• B. $\displaystyle \frac{a^{3}}{(b-c)(b-a)}$
• C. 1
• D. $\displaystyle \frac{a^{3}}{(a-b)(a-c)}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int \dfrac {\sin x \cos x}{\sqrt {1 - \sin^{4}x}}dx =$
• A. $\dfrac {1}{2}\,\cos^{-1} (\sin^{2}x) + C$
• B. $\tan^{-1}(\sin^{2} x) + C$
• C. $\tan^{-1}(2\sin x) + C$
• D. $\dfrac {1}{2}\, \sin^{-1}(\sin^{2}x) + C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate
$\int x^{2} \tan^{-1} xdx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve:
$\displaystyle \int 3^{3^{x^{x}}}3^{3^{x}}3^{x}dx=$

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