Evaluate the following integrals: ( + )^2dx - Vidyadip

Question

Evaluate the following integrals:
$\int(\tan\text{x}+\cot\text{x})^2\text{dx}$

✓

Answer

$\int(\tan\text{x}+\cot\text{x})^2\text{dx}$
$=\int(\tan^2\text{x}+\cot^2\text{x}+2\tan\text{x}\cot\text{x})\text{dx}$
$=\int(\tan^2\text{x}+\cot^2\text{x}+2)\text{dx}$
$=\int\big[(\sec^2\text{x}-1)+(\text{cosec}^2\text{x}-1)+2\big]\text{dx}$
$=\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "3 Marks" from INTEGRALS→INTEGRALS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Evaluate the following definite integrals:
$\int_{1}^\limits{\text{e}}\frac{\log\text{x}}{\text{x}}\text{ dx}$

Ten eggs are drawn successively, with replacement, from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.

Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.

Find the cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by $\frac{\text{x}+3}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+8}{6}.$

Find the principal values of the following:
$\tan^{-1}\Big(-\frac{1}{\sqrt3}\Big)$

Express $\overrightarrow{\text{AB}}$ in terms of unit vectors $\hat{\text{i}}\text{ and }\hat{\text{j}}$, when the point is:

A(4, -1), B(1, 3)

Find $\Big|\overrightarrow{\text{AB}}\Big|$

Find the second order derivatives of the following functions:
$\text{y}=\log(\log\text{x})$

Solve : $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$.

Find the inverse of the following matrices:

$\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}$

Give an example of a relation which is,
Reflexive and symmetric but not transitive.