((♭o♭hans)/(∼∼∼∼∼)) ∫_1 ^2 x sec^(−1) (x)dx=?


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Question Number 109097 by bobhans last updated on 21/Aug/20

$\frac{♭ o ♭ h a n s}{\sim\sim\sim\sim\sim}$ $\underset{1}{\int^{2}} x \sec^{- 1} (x) d x = ?$
	Answered by john santu last updated on 21/Aug/20
	$((⊸JS⊸)/(∤…∤…∤)) I=∫_1 ^2 x sec^(−1) (x) dx → { ((sec^(−1) (x)=k, x=sec k)),((x=1,k=0, x=2,k=(π/3))) :} I=∫_0 ^(π/3) k. sec k (sec k.tan k dk ) I=∫_0 ^(π/3) k.sec^2 k.tan k dk [ by parts → { ((u=k)),((v=∫tan k d(tan k)=(1/2)tan^2 k)) :}] I=(1/2)k.tan^2 k ]_0 ^(π/3) −(1/2)∫tan^2 k dk I=(1/2).(π/3).((√3))^2 −(1/2)∫ (sec^2 k−1)dk I=(π/2)−(1/2)[tan k−k ]_0 ^(π/3) I=(π/2)−(1/2)[((√3)−(π/3))−(0)] I=(π/2)−((√3)/2)+(π/6) = ((2π)/3)−((√3)/2).$
	$\frac{⊸ J S ⊸}{∤ \dots ∤ \dots ∤}$ $I = \underset{1}{\int^{2}} x \sec^{- 1} (x) d x \to {\begin{cases} \sec^{- 1} (x) = k, x = \sec k \\ x = 1, k = 0, x = 2, k = \frac{π}{3} \end{cases}$ $I = \underset{0}{\int^{π / 3}} k . \sec k (\sec k . \tan k d k)$ $I = \underset{0}{\int^{π / 3}} k . \sec^{2} k . \tan k d k$ $[b y p a r t s \to {\begin{cases} u = k \\ v = \int \tan k d (\tan k) = \frac{1}{2} \tan^{2} k \end{cases}]$ ${I = \frac{1}{2} k . \tan^{2} k]}_{0}^{π / 3} - \frac{1}{2} \int \tan^{2} k d k$ $I = \frac{1}{2} . \frac{π}{3} . {(\sqrt{3})}^{2} - \frac{1}{2} \int (\sec^{2} k - 1) d k$ $I = \frac{π}{2} - \frac{1}{2} {[\tan k - k]}_{0}^{π / 3}$ $I = \frac{π}{2} - \frac{1}{2} [(\sqrt{3} - \frac{π}{3}) - (0)]$ $I = \frac{π}{2} - \frac{\sqrt{3}}{2} + \frac{π}{6} = \frac{2 π}{3} - \frac{\sqrt{3}}{2} .$
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