Question

${\int }_{\sin \theta }^{\cos \theta }f\left(x\tan \theta \right)dx(where\theta \ne \frac{n\pi }{2},nϵI)$ is equal to

• A. $-{\int }_1^{\tan }f\left(x\sin \theta \right)dx$
• B. $-\tan \theta {\int }_{\sin \theta }^{\cos \theta }f\left(x\right)dx$
• C. $\sin \theta {\int }_1^{\tan }f\left(x\cos \theta \right)dx$
• D. $\frac{1}{\tan \theta }{\int }_{\sin \theta }^{\sin \theta \tan \theta }f\left(x\right)dx$

Answer 1

The correct option is A $-{\int }_1^{\tan }f\left(x\sin \theta \right)dx$

Let $I={\int }_{sin\theta }^{cos\theta }f\left(xtan\theta \right)dx$
Substitute $xtan\theta =zsin\theta \Rightarrow dx=cos\theta dz$
Therefore $x=sin\theta \Rightarrow z=tan\theta$ and $x=cos\theta \Rightarrow z=1$
Gives
$I={\int }_{tan\theta }^{1}f\left(zsin\theta \right)dz=-{\int }_1^{tan\theta }f\left(zsin\theta \right)dz=-{\int }_1^{tan\theta }f\left(xsin\theta \right)dx$