Browse Questions

# $g(t) =\int \limits_o^t \sin ^6 x .dx$ find $g(\pi+t)$

$\begin {array} {1 1} (a)\;g(t) +g(x) \\ (b)\;g(t)-g(x) \\ (c)\;\frac{g(t) +g(x)}{2} \\ (d)\;None \end {array}$

$g(\pi+t) =\int \limits_o^{\pi+t} \sin ^6 x .dx$
$g(\pi+t) =\int \limits_o^t \sin ^6 x .dx+\int \limits_{t+0}^{t+\pi} \sin ^6 x .dx$
$g(\pi+t) =g(t)+\int \limits_o^{\pi} \sin ^6 x .dx$
$g(\pi+t)=g(t) +g(x)$
Hence a is the correct answer.