265. ∫0π4·2tan3x dx=1−log2

Let I ∫0π4·2tan3x dx =2∫0π4tan2xtanx dx =2∫0π4(sec2x−1)tanx dx {?sec2x=tan2x+1} =2∫0π4sec2xtanx dx−2∫0π4tan x dx =2I1+2[log](cosx)0π4 =2I1+2[log(cosπ4)−log(cos0)] =2I1+2(log1/√2 −log1) =2I1+2(log·2−12−0) I=2I1+2*(−12)log2=2I1−log2.____(1). Where I 1=∫0π4sec2xtanx dx Let tan x = t =>sec2xdx = dt When, x = 0, t = tan 0. = 0 x=π4,t=tanπ4=1 1=∫01t dt=[t22]01=12−0=12 So, Equation (1) becomes, =2*12−log2 =1 - log 2.

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

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Vishal Baghel | Contributor-Level 10

7 months ago

Let I $\int_{0}^{\frac{π}{4}} \cdot 2 t a n^{3} x d x$

$= 2 \int_{0}^{\frac{π}{4}} t a n^{2} x t a n x d x$

$= 2 \int_{0}^{\frac{π}{4}} (s e c^{2} x - 1) t a n x d x {? s e c^{2} x = t a n^{2} x + 1}$

$= 2 \int_{0}^{\frac{π}{4}} s e c^{2} x t a n x d x - 2 \int_{0}^{\frac{π}{4}} t a n x d x$

$= 2I_{1} + 2 [l o g] {(c o s x)}_{0}^{\frac{π}{4}}$

$= 2I_{1} + 2 [l o g (c o s \frac{π}{4}) - l o g (c o s 0)]$

$= 2I_{1} + 2 (l o g1/√2$
$\frac{}{} - l o g 1)$

$= 2I_{1} + 2 (l o g \cdot 2^{\frac{- 1}{2}} - 0)$

$I = 2I_{1} + 2 * (\frac{- 1}{2}) l o g 2 = 2I_{1} - l o g 2 .____(1) .$

Where I $_{1} = \int_{0}^{\frac{π}{4}} s e c^{2} x t a n x d x$

Let tan x = t =>sec²xdx = dt

When, x = 0, t = tan 0. = 0

$x = \frac{π}{4}, t = t a n \frac{π}{4} = 1$

$_{1} = \int_{0}^{1} t d t = {[\frac{t^{2}}{2}]}_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}$

So, Equation (1) becomes,

$= 2 * \frac{1}{2} - l o g 2$

=1 - log 2.

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