Question

integrate cos^4 x dx ​

Answer 1

Topic :-

Indefinite Integration

To Integrate :-

$\displaystyle \int\cos^4x\:dx$

$\displaystyle \int(\cos^2x)^2\:dx$

$\displaystyle \int\left(\dfrac{1+\cos 2x}{2} \right)^2\:dx$

$\left(\because \cos^2x=\dfrac{1+\cos 2x}{2} \right)$

$\displaystyle \int\dfrac{(1+\cos 2x)^2}{4}\:dx$

$\displaystyle \int\dfrac{1+2\cos2x+\cos^22x}{4}\:dx$

$\displaystyle \int\dfrac{1}{4}\:dx+\int\dfrac{2\cos2x}{4}\:dx+\int\dfrac{\cos^22x}{4}\:dx$

$\displaystyle \dfrac{1}{4} \int dx+\dfrac{1}{2}\int\cos2x\:dx+\dfrac{1}{4}\int\cos^22x\:dx$

$\displaystyle \dfrac{1}{4} \int dx+\dfrac{1}{2}\int\cos2x\:dx+\dfrac{1}{4}\int\dfrac{1+\cos4x}{2}\:dx$

$\left(\because \cos^22x=\dfrac{1+\cos 4x}{2} \right)$

$\displaystyle \dfrac{1}{4} \int dx+\dfrac{1}{2}\int\cos2x\:dx+\dfrac{1}{8}\int dx+\dfrac{1}{8}\int\cos4x\:dx$

$\displaystyle \dfrac{1}{4}\cdot x+\dfrac{1}{2}\int\cos2x\:dx+\dfrac{1}{8}\cdot x+\dfrac{1}{8}\int\cos4x\:dx$

$\displaystyle \left( \because \int dz=z+C,where\:C\:is\:constant\:of\:integration.\right)$

$\displaystyle \dfrac{1}{4}\cdot x+\dfrac{1}{2}\cdot\dfrac{\sin 2x}{2}+\dfrac{1}{8}\cdot x+\dfrac{1}{8}\cdot\dfrac{\sin4x}{4}+C$

$\displaystyle \left( \because \int \cos az\:dz=\dfrac{\sin az}{a}+C,where\:C\:is\:constant\:of\:integration.\right)$

$\dfrac{x}{4}+\dfrac{\sin 2x}{4}+\dfrac{x}{8}+\dfrac{\sin4x}{32}+C$

$\dfrac{8x+8\sin 2x+4x+\sin4x}{32}+C$

$\dfrac{12x+8\sin 2x+\sin4x}{32}+C$

Answer :-

$\underline{\boxed{\displaystyle \int\cos^4x\:dx=\dfrac{12x+8\sin 2x+\sin4x}{32}+C}}$