IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

sunipduttagupta

Q. _{[ ]}

^{[ ]} sin 2x/(a² sin² x + b² cos² x)

rhd92781

well, this can be written as _{[ ]}

^{[ ]} sin2x/((a^2-b^2)sin²x + b^2)) dx

now put sin²x = t

u will get sin2x dx = dt

so it becomes

_{[ ]}

^{[ ]} dt/((a^2-b^2)t^2 + b^2)

= 1/(a^2-b^2) _{[ ]}

^{[ ]} dt/(t² + b/^{[ ]}

(a^2-b^2))

=1/b^{[ ]}

(a²-b²) tan^-1(t^{[ ]}

(a²-b²)/b) + c

now put t= sin²x in this and get the answer

Please rate !

krishna.gopal

Perfect answer rhd92781. As a practice sunip you can try same problem but by taking cos^2(x) as t

vidyun_dusa

divide numerator &denominator wit cos²x & then make da substitution tan²x=t