Evaluate the following integrals: ^ 2 _0 2x ^ -1 ( )dx - Vidyadip

Question

Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$

✓

Answer

Let $\text{I}=\int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$ Differentiating w.r.t. x, we get$\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0,\text{t}=0$$\text{x}=\frac{\pi}{2},\text{t}=1$
$\therefore\ \int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}=2\int^\limits{1}_0\text{t }\tan^{-1}\text{t dt}$ $\big[\because\sin2\text{x}=2\sin\text{x }\cos\text{x}\big]$
Using by parts$=2\Big\{\tan^{-1}\text{t}\int\text{t dt}-\int\big(\int\text{t dt}\big)\frac{\text{d}\tan^{-1}\text{t}}{\text{dt}}\Big\}$
$=2\Big\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\int\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}\Big\}$
$=2\bigg\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\Big(\int\text{dt}-\int\frac{\text{dt}}{1+\text{t}^2}\text{ dt}\Big)\bigg\}$
$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\big(\text{t}-\tan^{-1}\text{t}\big)\Big]^1_0$
$=2\bigg\{\frac{1}{2}\frac{\pi}{4}-\frac{1}{2}\Big(1-\frac{\pi}{4}\Big)\bigg\}$
$=2\Big\{\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}\Big\}$
$=2\Big(\frac{\pi}{4}-\frac{1}{2}\Big)$
$=\frac{\pi}{2}-1$

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

To maintain one's health, a person must fulfil certain minimum daily requirements for the following three nutrients: calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:

	Food I	Food II	Minimum daily requirement
Calcium	10	4	20
Protein	5	6	20
Calories	2	6	12
Price	Rs. 0.60 per unit	Rs. 1.00 per unit

Find the combination of food items so that the cost may be minimum.

The function $y=a \log x+b x^2+x$ has extreme values at $x=1$ and $x=2$. Find $a$ and $b$.

Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}=\theta+\sin\theta,\text{y}=1+\cos\theta\text{ at }\theta=\frac{\pi}{2}$

Use product $\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$ to solve the system of equations $x + 3z = 9, -x + 2y - 2z = 4, 2x - 3y + 4z = -3.$

Find the area of the region enclosed between the two curve $x^2 + y^2 = 9$ and $(x - 3)^2 + y^2 = 9.$

Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$

Find the point on the curve $y^2= 4x$ which is nearest to the point $(2, -8).$

Draw a rough sketch of the graph of the function $\text{y}=2\sqrt{1-\text{x}^{2}}, \text{x}\in [0, 1] $ and evaluate the area enclosed between the curve and the x-axis.

If A and B are two events such that,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ then find $\text{P}(\text{A}\cap\text{B}),$ P(A|B) and P(B|A).

find the area of the region bound by the curve $x = at ^2, y =2$ at between the ordinatrs corresponding $t =1$ and $t =2$.