Download App

Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx},0<\alpha<\pi$
✓

Answer

We have,$\text{I}=\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get,$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}+\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{\sec^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$$\frac{1}{2}\sec^2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{t}\rightarrow\infty$$\therefore\ \text{I}=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{1+\text{t}^2+2\cos\alpha\text{t}}\text{ dt}$
$=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{(\text{t}+\cos\alpha)^2-\cos^2\alpha+1}\text{ dt}$
$={\pi}\int\limits^{\infty}_0\frac{1}{(\text{t}+\cos\alpha)^2+\sin^2\alpha}\text{ dt}$
$=\pi\bigg[\frac{1}{\sin\alpha}\tan^{-1}\Big(\frac{1+\cos\alpha}{\sin\alpha}\Big)\bigg]^1_0$
$=\frac{\pi}{\sin\alpha}\Big[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\Big]$
$=\frac{\pi}{\sin\alpha}\bigg[\frac{\pi}{2}-\tan^{-1}\Big(\tan\Big(\frac{\pi}{2}-\alpha\Big)\Big)\bigg]$
$=\frac{\pi\alpha}{\sin\alpha}$

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 Definite Integrals→Definite Integrals — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Differentiate $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$ with respect to $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
→
Find the area of the ragion in the first quadrant bounded by the parabola $y=4 x^2$ and the lines $x=0, y=1$ and $y=4$.
→
Water is running into an inverted cone at the rate of $\pi$ cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5m. How fast the water level is rising when the water stands 7.5m below the base.
→
Prove that the medians of a triangle are concurrent.
→
Five defective mangoes are acciedently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.
→
If $\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0,$ then using properties of determinants, find the value of $\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}},$ where $\text{x},\text{y},\text{z}\neq0.$
→
If $\text{xy}=4,$ prove that $\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
→
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}-\text{y}=\text{xe}^{\text{x}}$
→
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 2x^2 - 24x + 7$
→
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}=-4\text{xy}^2$ given that $\text{y}=1.$ when $\text{x}=0.$
→