Download App

INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following intregals:
$\int\frac{1}{13+3\cos\text{x}+4\sin\text{x}}\ \text{dx}$
✓

Answer

Let $\text{I}=\int\frac{1}{13+3\cos\text{x}+4\sin\text{x}}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\sin\text{x}=\frac{2\tan\Big(\frac{\text{x}}{2}\Big)}{1+\tan^2\Big(\frac{\text{x}}{2}\Big)}$
$\therefore\text{I}=\int\frac{1}{13+3\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+4\times2\frac{\tan\Big(\frac{\text{x}}{2}\Big)}{1+\tan^2\Big(\frac{\text{x}}{2}\Big)}}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{13\Big(1+\tan^2\frac{\text{x}}{2}\Big)+3-3\tan^2\frac{\text{x}}{2}+16+8\tan\Big(\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{13\tan^2\frac{\text{x}}{2}-3\tan^2\frac{\text{x}}{2}+16+8\tan\Big(\frac{\text{x}}{2}\Big)}\ \text{dx}$
$==\int\frac{\sec^2\frac{\text{x}}{2}}{10\tan^2\Big(\frac{\text{x}}{2}\Big)+8\tan\Big(\frac{\text{x}}{2}\Big)+16}\ \text{dx}$
Let $\tan\Big(\frac{\text{x}}{2}\Big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\text{I}=\int\frac{2\text{dt}}{10\text{t}^2+8\text{t}+16}$
$=\int\frac{\text{dt}}{5\text{t}^2+4\text{t}+8}$
$=\frac{1}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{4}{5}\text{t}+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{4}{5}\text{t}+\Big(\frac{2}{5}\Big)^2-\Big(\frac{2}{5}\Big)^2+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)^2-\frac{4}{25}+\frac{8}{5}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)^2+\frac{-4+40}{25}}$
$=\frac{1}{5}\int\frac{\text{dt}}{\Big(\text{t}+\frac{2}{5}\Big)+\Big(\frac{6}{5}\Big)^2}$
$=\frac{1}{5}\times\frac{5}{6}\tan^{-1}\Bigg(\frac{\text{t}+\frac{2}{5}}{\frac{6}{5}}\Bigg)+\text{C}$
$=\frac{1}{6}\tan^{-1}\Big(\frac{5\text{t}+2}{6}\Big)+\text{C}$
$=\frac{1}{6}\tan^{-1}\bigg(\frac{5\tan\frac{\text{x}}{2}+2}{6}\bigg)+\text{C}$

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 "4 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

If $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
→
Maximise and Minimise Z = 3x – 4y subject to:
$\text{x}-2\text{y}\leq0$
$-3\text{x}+\text{y}\leq4$
$\text{x}-\text{y}\leq6$
$\text{x},\text{y}\geq0$
→
Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
by logarithmic differentiation.
→
Evaluate the following integrals:

$\int\text{e}^{\text{x}}\Big(\frac{\sin4\text{x}-4}{1-\cos4\text{x}}\Big)\text{dx}$

→
Solve the following differential equations:
$2(\text{y}+3)-\text{xy}\frac{\text{dy}}{\text{dx}}=0,\text{y}(1)=-2$
→
Prove that $\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$ is divisible by (a + b + c) and find the quotient.
→
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}} = (\text{x}+\text{y})^2$
→
Evaluate the following integrals:

$\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$

→
Evaluate the following integrals:
$\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
→
If $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big),$ prove that $\big(\text{x}^2+1\big)\frac{\text{dx}}{\text{dx}}+\text{xy}+1=0$
→