Download App

Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{1}{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}\ \text{dx}$
✓

Answer

Consider the integral
$\text{I}=\int\frac{1}{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}\ \text{dx}$
Divide both the numerator and the denominator by $\cos^4\text{x}$, we have,
$\text{I}=\int\frac{\frac{1}{\cos^4\text{x}}}{\frac{\sin^4\text{x}+\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}}{\cos^4\text{x}}}\ \text{dx}$
$=\int\frac{\sec^4\text{x}}{\tan^4\text{x}+\tan^2\text{x}+1}\ \text{dx}$
$=\int\frac{\sec^2\text{x}\times\sec^2\text{x}}{\tan^4\text{x}+\tan^2\text{x}+1}\ \text{dx}$
$=\int\frac{(\tan^2\text{x}+1)\text{x}\sec^2\text{x}}{\tan^4\text{x}+\tan^2\text{x}+1}\ \text{dx}$
Substituting $\tan\text{x}=\text{t};\ \sec^2\text{x dx}=\text{dt}$
Thus
$\text{I}=\int\frac{(1+\text{t})^2}{\text{t}^4+\text{t}^2+1}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}^2+\frac{1}{\text{t}^2}+1\Big)}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}^2+\frac{1}{\text{t}^2}-2+2+1\Big)}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}-\frac{1}{\text{t}}\Big)^2+3}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}-\frac{1}{\text{t}}\Big)^2+3}$
Substituting $\text{z}=\text{t}=-\frac{1}{\text{t}};\text{dz}=\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}$
$\text{I}=\int\frac{\text{dz}}{\text{z}^2+3}$
$\Rightarrow\text{I}=\int\frac{\text{dz}}{\text{z}^2+(\sqrt{3})^2}$
$\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{z}}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{t}-\frac{1}{\text{t}}}{\sqrt{3}}\Bigg)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\bigg(\frac{\tan\text{x}-\frac{1}{\tan\text{x}}}{\sqrt{3}}\bigg)+\text{C}$
$\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\tan\text{x}-\cot\text{x}}{\sqrt{3}}\Big)+\text{C}$

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 "Solve the Following Question.(5 Marks)" from Indefinite Integrals→Indefinite Integrals — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following functions. Also, find the points of inflection,$f'(x) = (x - 1)(x + 2)^2$
→
Differentiate the following functions from first principles:
$\text{e}^{\cos\text{x}}$
→
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 and twice of its y-intercept is equal to three times its z-intercept.
→
Evaluate the following intregals:$\int\frac{1}{3+2\cos^2\text{x}}\text{ dx}$
→
Evaluate : $\int_0^2 \frac{2^x}{2^x\left(1+4^x\right)} \cdot d x$
→
Evaluate the following intregals:
$\int\frac{1}{5+7\cos\text{x}+\sin\text{x}}\ \text{dx}$
→
For the following matrices verify the associativity of multiplication i.e., (AB) C = A(BC):
$\text{A}=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&0\\-1&2\\0&3\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1\\-1\end{bmatrix}$
→
Find the equation of the normal to the curve $x^2 + 2y^2 - 4x - 6y + 8 = 0$ at the point whose abscissa is $2.$
→
Find the equation of the plane which contains planes is the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0 and whose x-intercept is twice its z-intercept.
→
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=​​\begin{cases}\sin\text{x}-\cos\text{x},&\text{if }\text{ x}\neq0\\-1,&\text{if }\text{ x}=0\end{cases}$
→