Show the steps needed to find the given solutions. ∫arctan⁡c1+x2dx

1) ∫arctan⁡(x)1+x2dx Let u=arctan⁡(x) dudx=ddx(arctan⁡(x)) dudx=11+x2⇒dx=du(1+x2) Substiute dx value in eq. 1 ⇒∫u⋅du(1+x2)(1+x2)=∫udu=u22+C Substitute "u" ⇒(arctan⁡(x))22+C

2) ∫cos2xsin⁡xdx Let u=cos⁡x⇒dudx=ddx(cos⁡x) =−sin⁡x Substitute dx "value inequality" du−sin⁡x=dx ∫u3sin⁡xdu−sin⁡x=∫−u2du=−u44+C Substitute "u" value =−cos4x4+C

3) ∫sec2⁡xtan3⁡xdx Let u=tan⁡x:⇒dudx=ddx(tan⁡x) dx=1sec2⁡xdu Substitute "dx" value in equation ∫sec2⁡xu2dusec2⁡x=u44+C=tan4⁡(x)4+C

1) ∫arctan⁡(x)1+x2dx Let u=arctan⁡(x) dudx=ddx(arctan⁡(x)) dudx=11+x2⇒dx=du(1+x2) Substiute dx value in eq. 1 ⇒∫u⋅du(1+x2)(1+x2)=∫udu=u22+C Substitute "u" ⇒(arctan⁡(x))22+C

2) ∫cos2xsin⁡xdx Let u=cos⁡x⇒dudx=ddx(cos⁡x) =−sin⁡x Substitute dx "value inequality" du−sin⁡x=dx ∫u3sin⁡xdu−sin⁡x=∫−u2du=−u44+C Substitute "u" value =−cos4x4+C

3) ∫sec2⁡xtan3⁡xdx Let u=tan⁡x:⇒dudx=ddx(tan⁡x) dx=1sec2⁡xdu Substitute "dx" value in equation ∫sec2⁡xu2dusec2⁡x=u44+C=tan4⁡(x)4+C

Step by step solution to "Show the steps needed to find the given..."

chezmarylou1i

Answered question

2021-12-23

Show the steps needed to find the given solutions.

\int \frac{\arctan c}{1 + x^{2}} d x

Answer & Explanation

Travis Hicks

Beginner2021-12-24Added 29 answers

\int \frac{\arctan (x)}{1 + x^{2}} d x

Let

u = \arctan (x)

\frac{d u}{d x} = \frac{d}{d x} (\arctan (x))

\frac{d u}{d x} = \frac{1}{1 + x^{2}} \Rightarrow d x = d u (1 + x^{2})

Substiute dx value in eq. 1

\Rightarrow \int \frac{u \cdot d u (1 + x^{2})}{(1 + x^{2})} = \int u d u = \frac{u^{2}}{2} + C

Substitute "u"

\Rightarrow \frac{{(\arctan (x))}^{2}}{2} + C

Jeffery Autrey

Beginner2021-12-25Added 35 answers

\int \cos^{2} x \sin x d x

Let

u = \cos x \Rightarrow \frac{d u}{d x} = \frac{d}{d x} (\cos x)

= - \sin x

Substitute dx "value inequality"

\frac{d u}{- \sin x} = d x

\int u^{3} \sin x \frac{d u}{- \sin x} = \int - u^{2} d u = \frac{- u^{4}}{4} + C

Substitute "u" value

= \frac{- \cos^{4} x}{4} + C

karton

Expert2021-12-30Added 613 answers

3) $\int \sec^{2} x \tan^{3} x d x$
Let $u = \tan x :\Rightarrow \frac{d u}{d x} = \frac{d}{d x} (\tan x)$
$d x = \frac{1}{\sec^{2} x} d u$
Substitute "dx" value in equation
$\int \sec^{2} x u^{2} \frac{d u}{\sec^{2} x} = \frac{u^{4}}{4} + C = \frac{\tan^{4} (x)}{4} + C$

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