Question

Evaluate
$integral$
cos⁴dx​

Answer 1

Question :

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

Solution :

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

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

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

$\bullet\ \; \sf \green{cos^2 \theta = \dfrac{1+cos\ 2 \theta}{2}}$

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

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

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

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

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

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

$\bullet\ \; \sf \red{\int cos\ x= sin\ x\ \; \&\ \; \int cos\ 2x = \frac{sin\ 2x}{2}}$

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

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

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

★ ═════════════════════ ★

Answer 2

Step-by-step explanation:

Question :

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

4

x dx

Solution :

\displaystyle \sf \int cos^4x\ dx∫cos

4

x dx

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

2

x)

2

dx

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

2

θ

\bullet\ \; \sf \green{cos^2 \theta = \dfrac{1+cos\ 2 \theta}{2}}∙ cos

2

θ=

2

1+cos 2θ

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

2

1+cos 2x

)

2

dx

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

4

1+cos

2

2x+2 cos 2x

dx

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

2

(2θ)=

2

1+cos 2(2θ)

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

4

1+

2

1+cos 2(2x)

+2 cos 2x

dx

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

8

2+1+cos 4x+4 cos 2x

dx

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

8

3+cos 4x+4 cos 2x

dx

\bullet\ \; \sf \red{\int cos\ x= sin\ x\ \; \&\ \; \int cos\ 2x = \frac{sin\ 2x}{2}}∙ ∫cos x=sin x & ∫cos 2x=

2

sin 2x

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

8

1

∫(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⟶

8

1

[3x+

4

cos 4x

+4

2

cos 2x

] +c

\displaystyle \longrightarrow \sf \pink{\dfrac{1}{8} \left[ 3x+ \dfrac{cos\ 4x}{4}+2\ cos\ 2x \right]\ + c}⟶

8

1

[3x+

4

cos 4x

+2 cos 2x] +c

★ ═════════════════════ ★

Thanks for your question.