Solve the trigonometric integral $\int\frac{\tan\left(x\right)^2\sec\left(x\right)^2}{\left(2+\tan\left(x\right)^3\right)^2}dx$

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Final answer to the problem

$\frac{1}{-3\left(2+\tan\left(x\right)^3\right)}+C_0$
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Step-by-step Solution

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  • Integrate by partial fractions
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We can solve the integral $\int\frac{\tan\left(x\right)^2\sec\left(x\right)^2}{\left(2+\tan\left(x\right)^3\right)^2}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2+\tan\left(x\right)^3$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

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$u=2+\tan\left(x\right)^3$

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Learn how to solve trigonometric integrals problems step by step online. Solve the trigonometric integral int((tan(x)^2sec(x)^2)/((2+tan(x)^3)^2))dx. We can solve the integral \int\frac{\tan\left(x\right)^2\sec\left(x\right)^2}{\left(2+\tan\left(x\right)^3\right)^2}dx by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it u), which when substituted makes the integral easier. We see that 2+\tan\left(x\right)^3 it's a good candidate for substitution. Let's define a variable u and assign it to the choosen part. Now, in order to rewrite dx in terms of du, we need to find the derivative of u. We need to calculate du, we can do that by deriving the equation above. Isolate dx in the previous equation. Substituting u and dx in the integral and simplify.

Final answer to the problem

$\frac{1}{-3\left(2+\tan\left(x\right)^3\right)}+C_0$

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Plotting: $\frac{1}{-3\left(2+\tan\left(x\right)^3\right)}+C_0$

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5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Trigonometric Integrals

Integrals that contain trigonometric functions and their powers.

Used Formulas

See formulas (2)

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