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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate: $\int \limits_{_z}^{z} \large\frac{2x(1+\sin x)}{1+\cos ^2 x}$$dx$

\[\begin {array} {1 1} (a)\;z^2 \\ (b)\;z^3 \\ (c)\;2z \\ (d)\;\frac{z^2}{4} \end {array}\]

1 Answer

$\int \limits_{_z}^{z} \large\frac{2x}{1+\cos ^2 x}+ \int \limits_{_z}^{z} \large\frac{2x \sin x)}{1+\cos ^2 x}$$dx$
By applying the property
$\int \limits_{-a} ^a f(x)dx=> 2 \int _0^a f(x). f(x) =f(-x)$
=>$ 0$$\qquad f(x) =-f(-x)$
$=0+ 2 \int \limits_0^z \large\frac{2x \sin x}{1+\cos ^2 x} dx$
By solving this
$\qquad=4 \times \large\frac{z^2}{4}$
$\qquad=z^2$
Hence a is the correct answer
answered Dec 13, 2013 by meena.p
 
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