$\int_{- π}^{π} \frac{2 x (1 + s i n x)}{1 + c o s^{2} x}$

Open in App

Solution

$= \int_{- π}^{π} \frac{2 x (1 + s i n x)}{1 + c o s^{2} x} d x \int_{- π}^{π} \frac{2 x)}{1 + c o s^{2} x} d x + \frac{2 x s i n x}{1 + c o s^{2} x} d x = \int_{- π}^{π} f_{1} (x) d x + \int_{- π}^{π} f_{2} (x) d x$
$f_{1} (x)$ is an odd function, because $f_{1} (- x) = - f_{1} (x)$
$∴ \int_{- π}^{π} f_{1} (x) d x = 0 f_{2} (- x) = \frac{2 (- x) . s i n (- x)}{1 + c o s^{2} (- x)} = \frac{2 x s i n x}{1 + c o s^{2} x} = f_{2} (x)$
$∴ f_{2} (x)$ is an even function
$\int_{- π}^{π} f_{1} (x) d x = 2 \int_{- π}^{π} f_{2} (x) d x$
$2 \int_{0}^{π} \frac{2 x s i n x}{1 + c o s^{x}} d)$
Put $c o s x = t s i n x d x = d t$
$x = 0 t = 1 x = π f = - 1 = 4 \int_{1}^{- 1} \frac{d t}{1 + t^{2}} = 4 \int_{1}^{- 1} \frac{d t}{1 + t^{2}} = - 4 [t a n^{- 1} t]^{2} = - 4 [\frac{π}{4} - (\frac{- π}{4})] = 4 \frac{π}{2} = 2 π ∴ I = \int_{π}^{π} \frac{2 x (1 + s i n x)}{1 + c o s^{2} x} d x = 2 π$

Suggest Corrections

Similar questions

π \int - π \frac{2 x (1 + sin x)}{1 + {cos}^{2} x} d x

\int_{- π}^{π} \frac{2 x (1 + sin x)}{1 + {cos}^{2} x} d x

\int_{- π}^{π} \frac{2 x (1 + s i n x)}{1 + c o s^{2} x} d x .

I = π \int - π \frac{2 x (1 + sin x) d x}{1 + {cos}^{2} x}

[\frac{I}{π^{2}}]

\frac{1 - sin x + \dots . . + (- 1)^{n} {sin}^{n} x .}{1 + sin x + . . . . + s i n^{n} x} = \frac{1 - cos 2 x}{1 + cos 2 x}

Join BYJU'S Learning Program