Prove the following question- -0-4 2 tan 3 x d x-1-log 2-

Let I=∫_0^π/42 tan^3x dx=2∫_0^π/4tan^2x.tan x dx =2∫_0^π/4(sec^2x 1)tan x dx [∵ 1+tan^2x=sec^2x] =2[∫_0^π/4sec^2x tan x dx ∫_0^π/4tan x dx] =2∫_0^π/4(tan ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

Prove the fol...

Question

Prove the following question.

$\int_{0}^{\frac{π}{4}} 2 t a n^{3} x d x = 1 - l o g 2.$

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Solution

$L e t I = \int_{0}^{\frac{π}{4}} 2 t a n^{3} x d x = 2 \int_{0}^{\frac{π}{4}} t a n^{2} x . t a n x d x = 2 \int_{0}^{\frac{π}{4}} (s e c^{2} x - 1) t a n x d x [∵ 1 + t a n^{2} x = s e c^{2} x] = 2 [\int_{0}^{\frac{π}{4}} s e c^{2} x t a n x d x - \int_{0}^{\frac{π}{4}} t a n x d x] = 2 \int_{0}^{\frac{π}{4}} (t a n x) s e c^{2} x d x - 2 {[- l o g | c o s x |]}_{0}^{\frac{π}{4}} [L e t, ∵ I_{1} = \int (t a n x) s e c^{2} x d x p u t t a n x = t \Rightarrow s e c^{2} x d x = d t ∴ I_{1} = \int t d t = \frac{t^{2}}{2} = \frac{t a n t^{2} x}{2}] = 2 {[\frac{t a n^{2} x}{2}]}_{0}^{\frac{π}{4}} + 2 [l o g ∣ ∣ c o s \frac{π}{4} ∣ ∣ - l o g | c o s 0 |] = t a n^{2} (\frac{π}{4}) - 0 + 2 l o g (\frac{1}{\sqrt{2}}) - l o g 1 = 1 + 2 l o g x^{- \frac{1}{2}} - 0 (∵ l o g 1 = 0) = 1 - 2 \times \frac{1}{2} l o g 2 = 1 - l o g 2. h e n c e p r o v e d .$

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Similar questions

Q. Prove

\int_{0}^{\frac{π}{4}} 2 {tan}^{3} x d x = 1 - log 2

Q. Prove that

\int_{0}^{π / 4} 2 {tan}^{3} x d x = 1 - log 2

Q. The value of

\frac{π}{4} \int 0 2 {tan}^{3} x d x

Prove the following question.

$\int_{0}^{\frac{π}{2}} s i n^{3} x d x = \frac{2}{3} .$

Prove the following question.

$\int_{1}^{3} \frac{1}{x^{2} (x + 1)} d x = \frac{2}{3} + l o g \frac{2}{3}$

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Prove the following question. ∫π402 tan3x dx=1−log 2.

Prove the following question.

$\int_{0}^{\frac{π}{4}} 2 t a n^{3} x d x = 1 - l o g 2.$