Evaluate the following definite integrals: _ 0 ^ 1 1-x 1+x dx - Vidyadip

Question

Evaluate the following definite integrals:
$\int\limits_{0}^{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$

✓

Answer

We have,
$\int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
Let $\text{x}=\cos2\theta$
$\text{dx}=-2\sin2\theta\text{ d}\theta$
Now,
$\int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
$=\int_{\frac{\pi}{4}}^\limits{0}\frac{1-\cos2\theta}{1+\cos2\theta}\times(-2\sin2\theta)\text{d}\theta$
$\int_{\frac{\pi}{4}}^\limits{0}\frac{2\sin^2\theta}{2\cos^2\theta}\times2\sin2\theta\text{ d}\theta$ $\bigg[\because\ -\int_{\text{a}}^\limits{\text{b}}\text{f(x)}\text{dx}=\int_{\text{b}}^\limits{\text{a}}\text{f(x)}\text{dx}\bigg]$
$=\int_{\frac{\pi}{4}}^\limits{0}\frac{4\sin^3\theta}{\cos\theta}\text{ d}\theta$
Let $\cos\theta=\text{t}$
$-\sin\theta\text{ d}\theta=\text{dt}$
Now,
$\theta=0\Rightarrow\text{t}=1$
$\theta=\frac{\pi}{4}\Rightarrow\text{t}=\frac{1}{\sqrt{2}}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{4}}\frac{4\sin^3\theta}{\cos\theta}\text{ d}\theta$
$=-4\int_{1}^\limits{\frac{1}{\sqrt{2}}}\frac{\big(1-\text{t}^2\big)}{\text{t}}\text{ dt}$
$=-4\Big[\log\text{t}-\frac{\text{t}^2}{2}\Big]^{\frac{1}{\sqrt{2}}}_1$
$=-4\Big[\log\Big(\frac{1}{\sqrt{2}}\Big)-\frac{1}{4}-0+\frac{1}{2}\Big]$
$=-4\Big[\log\sqrt{2}+\frac{1}{4}\Big]$
$\therefore\ \int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}=2\log2-1$

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 "5 Marks Questions" from Definite Integrals→Definite 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 $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$, then find the value of $\frac{\text{f}^{2}\text{y}}{\text{dx}}\text{at}\theta = \frac{\pi}{6}.$

Examine the continuity of the function
$\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0$
Also sketch the graph of this function.

If $\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}},$ prove that $\frac{\text{y}^2\cot\text{x}}{(1-\text{y}\log\sin\text{x})}$

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{{27}}{\pi}h^3\tan^2 \alpha .$

Differentiate the following functions with respect to x:
$\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}$

Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\cos^{2}\text{x}}{1+\text{e}^{\text{x}}}\text{ dx}$

$\int\text{x}(1-\text{x})^{23}\text{dx}$

A diet of two foods $F_1$ and $F_2$ contains nutrients thiamine, phosphorous and iron.
The amount of each nutrient in each of the food (in milligrams per 25gms) is given in the following table:

Nutrients	Food	$F_1$	$F_2$
Thiamine		0.25	0.10
Phosphorous		0.75	1.50
Iron		1.60	0.80

The minimum requirement of the nutrients in the diet are 1.00mg of thiamine, 7.50mg of phosphorous and 10.00mg of iron.
The cost of $F_1$ is 20 paise per 25gms while the cost of $F_2$ is 15 paise per 25gms.
Find the minimum cost of diet.

Evaluvate the following intregals:
$\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}$

Show that the matrix $\text{A}=\begin{bmatrix}5&3\\12&7\end{bmatrix}$ is root of the equation $A^2 - 12A - I = 0$.