Question

sin3x cos4x .dx
Integrate the function

Answer 1

HELLO DEAR,

GIVEN function is integral of (sin3xcos4x).dx

we know sinAcosB = $\sf{\frac{1}{2}[sin(A + B)+sin(A - B)]}$

therefore, integral of (sin3xcos4x).dx

= $\sf{\int{}\frac {1}{2}[sin7x+sin(-x)]dx}$

= $\sf{\int{}\frac{1}{2}[sin7x-sinx]dx}$

= $\sf{\int{\frac{1}{2}sin7x}\,dx- \int{\frac{1}{2} sinx}\,dx}$

= $\sf{\frac{-cos7x}{14}+\frac{cos}{2}+c}$

I HOPE ITS HELP YOU DEAR,
THANK