Question

Q1.Evaluate
integral
cos⁴dx​

Answer 1

Step-by-step explanation:

Question :

$\ast\ \; \displaystyle \sf \red{Evaluate\ \int cos^4 x\ dx}∗$

Solution :

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

$\displaystyle \longrightarrow \sf \int (cos^2x)^2\ dx$

$\bullet\ \; \sf \orange{1+cos\ 2 \theta = 2\ cos^2 \theta}∙$

$\bullet\ \; \sf \green{cos^2 \theta = \dfrac{1+cos\ 2 \theta}{2}}∙ </p><p> </p><p>$

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

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

$\bullet\ \; \sf \purple{cos^2(2 \theta)=\dfrac{1+cos\ 2(2 \theta)}{2}}∙$

$\displaystyle \longrightarrow \sf \int \dfrac{1+\frac{1+cos\ 2(2x)}{2}+2\ cos\ 2x}{4}\ dx$

$\displaystyle \longrightarrow \sf \int \dfrac{2+1+cos\ 4x+4\ cos\ 2x}{8}\ dx$

$\displaystyle \longrightarrow \sf \int \dfrac{3+cos\ 4x+4\ cos\ 2x}{8}\ dx$

$\bullet\ \; \sf \red{\int cos\ x= sin\ x\ \; \&\ \; \int cos\ 2x = \frac{sin\ 2x}{2}}∙ </p><p>$

$\displaystyle \longrightarrow \sf \dfrac{1}{8} \int (3+cos\ 4x+4\ cos\ 2x)\ dx$

$\displaystyle \longrightarrow \sf \dfrac{1}{8} \left[ 3x+ \dfrac{cos\ 4x}{4}+4 \dfrac{cos\ 2x}{2} \right]\ + c</p><p>$

$\displaystyle \longrightarrow \sf \pink{\dfrac{1}{8} \left[ 3x+ \dfrac{cos\ 4x}{4}+2\ cos\ 2x \right]\ + c} </p><p>$

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Answer 2

Answer:

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