Question

Integrate the following
$\displaystyle{ \sf \: l = \int \: {tan}^{4} x.dx }$
From the limits [0,pi/4]​

Answer 1

Solution:

I am solving this integration first by Indefinite Integration and applying the limits in the later stage.

Step 1: Convert Tan⁴x as Tan²x . Tan²x

→ I = ∫Tan⁴x.dx = ∫(Tan²x.Tan²x)dx

According to identities,

→ 1 + Tan²x = Sec²x

→ Tan²x = Sec²x - 1

Step 2: Convert one Tan²x based on the above identity.

→ ∫(Tan²x .Tan²x)dx = ∫(Sec²x - 1).Tan²x. dx

Step 3: Multiplying the terms inside, we get:

→ ∫(Sec²x.Tan²x - Tan²x).dx

Step 4: Let us substitute Tan x = t

→ Sec²x.dx = dt

→ ∫(Sec²x.Tan²x - Tan²x).dx = ∫(Sec²x.Tan²x)dx  - ∫(Tan²x).dx

→ ∫(Sec²x.Tan²x)dx  - ∫(Tan²x).dx = ∫t².dt - ∫(Tan²x).dx

Solving ∫t².dt, we get:

→ [ t³/3 ] + C

Substituting t = Tan x, we get:

→ [ t³/3 ] + C = Tan³x/3 + C    → I₁

Now solving -∫(Tan²x).dx, we get:

→ -∫(Sec²x - 1).dx   [Applying Identity]

→ -∫Sec²x.dx - 1∫dx

→ -Tan x + x + C  [Integral Sec²x = Tan x]   → I₂

Combining both we get:

→ ∫Tan⁴x.dx = I₁ + I₂

→  ∫Tan⁴x.dx = Tan³x/3 + C - Tan x + x + C  

[Two constants add to give a common constant C]

→ ∫Tan⁴x.dx = Tan³x/3 - Tan x + x + C  

Now applying Limits, we get:

Applying Upper Limit: π/4

→ [Tan (π/4)]³ / 3 - Tan (π/4) + (π/4)

→ ( 1 )³ / 3 - 1 + π/4

→ 1/3 - 1 + π/4

→ π/4 - 2/3

Applying Lower Limit: 0

→ [Tan (0)]³ / 3 + Tan (0) - (0)

→ 0/3 + 0 - 0 = 0

Hence the final result is: π/4 - 2/3 - 0 = π/4 - 2/3

Therefore the value of integral after applying limits is: (π/4 - 2/3)

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

Integrate the following : $\sf{\int _0^{\tfrac{\pi }{4}}\:tan^4\left(x\right)\cdot dx}$

★═════════════════★  

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{-\frac{2}{3}+\frac{\pi }{4}}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\int _0^{\tfrac{\pi }{4}}\:tan^4\left(x\right)\cdot dx}$

Apply Integral Reduction :

$\sf{\int \tan ^{n}(x) d x=\dfrac{\tan ^{n-1}(x)}{n-1}-\int \tan ^{n-2}(x) d x}$

$\sf{\int \tan ^4\left(x\right)dx=\dfrac{\tan ^3\left(x\right)}{3}-\int \tan ^2\left(x\right)dx}$

$\sf{=\left[\dfrac{\tan ^3\left(x\right)}{3}\right]^{\tfrac{\pi }{4}}_0-\int _0^{\tfrac{\pi }{4}}\tan ^2\left(x\right)dx}$

$\bf{\int _0^{\tfrac{\pi }{4}}\tan ^2\left(x\right)dx\::}$

$\begin{aligned}&\text { Use the following identity: } \sec ^{2}(x)-\tan ^{2}(x)=1\\&\text { Therefore } \tan ^{2}(x)=-1+\sec ^{2}(x)\end{aligned}$

$\sf{=\int _0^{\tfrac{\pi }{4}}-1+\sec ^2\left(x\right)dx}$

$\begin{aligned}&\text { Apply the Sum Rule: } \int f(x) \pm g(x) d x=\int f(x) d x \pm \int g(x) d x\\&=-\int_{0}^{\frac{\pi}{4}} 1 d x+\int_{0}^{\frac{\pi}{4}} \sec ^{2}(x) d x\end{aligned}$

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$\begin{array}{l}\bf{\int_{0}^{\tfrac{\pi}{4}} 1 d x=\dfrac{\pi}{4} }\\\\\bf{\int_{0}^{\tfrac{\pi}{4}} \sec ^{2}(x) d x=1}\end{array}$

$=-\dfrac{\pi }{4}+1$

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$\sf{=\left[\dfrac{\tan ^3\left(x\right)}{3}\right]^{\tfrac{\pi }{4}}_0-\left(-\dfrac{\pi }{4}+1\right)}$

$\sf{\displaystyle=\left[\frac{\tan ^3\left(x\right)}{3}\right]^{\frac{\pi }{4}}_0+\frac{\pi }{4}-1}$

$\sf{\displaystyle=\left[\frac{1}{3}\tan ^3\left(x\right)\right]^{\frac{\pi }{4}}_0+\frac{\pi }{4}-1}$

Compute the boundaries : $\left[\dfrac{1}{3} \tan ^{3}(x)\right]_{0}^{\tfrac{\pi}{4}}=\dfrac{1}{3}$

$\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)$

$\begin{array}{l}\bf{\lim _{x \rightarrow 0+}\left(\dfrac{1}{3} \tan ^{3}(x)\right)=0} \\\\\bf{\lim _{x \rightarrow \frac{\pi}{4}-\left(\tfrac{1}{3} \tan ^{3}(x)\right)}=\dfrac{1}{3}}\end{array}$

$=\dfrac{1}{3}-0$

$=\dfrac{1}{3}$

$\boxed{\sf{=-\frac{2}{3}+\frac{\pi }{4}}}$