Question

Integrate the following function:
\[\int{\sin 4x.\cos 3xdx}\]

Answer 1

Hint: Firstly Multiply and divide with 2 in the numerator and denominator respectively then we will get numerator in terms of \[2\sin A\cos B\] then split the numerator using trigonometric formula given by \[2\sin A\cos B=\sin (A+B)+\sin (A-B)\] then obtain the integrals of each term using their respective formulas.

Complete step-by-step answer:
To find the \[\int{\sin 4x.\cos 3xdx}\]
Multiply and divide with 2 in the numerator and denominator respectively then we will get,
\[=\dfrac{1}{2}\int{2\sin 4x.\cos 3xdx}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know that \[2\sin A\cos B=\sin (A+B)+\sin (A-B)\]
\[=\dfrac{1}{2}\int{\left( \sin 7x+\sin x \right)dx}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
\[=\dfrac{1}{2}\left[ \dfrac{-\cos 7x}{7}-\cos x \right]+c\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
\[=-\dfrac{\cos 7x}{14}-\dfrac{\cos x}{2}+c\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
Note: The integral value of \[\sin x\] is given by \[-\cos x\] and the integral value of \[\sin ax\] is given by \[\dfrac{-\cos ax}{a}\]. For this type of problems we should know the basic trigonometric formulas, identities etc and integrals of the basic trigonometric functions. The constant of integration “c” is used for indefinite integrals.