It's another way!

I=∫xdx(x2+1)=12∫dx21+x2I=\int\dfrac{\red{xdx}}{(x^2+1)}={\dfrac{1}{2}}\int\dfrac{d\red{x^2}}{1+x^2} =12∫d(x2+1)x2+1=\dfrac{1}{2}\int\dfrac{d(\red{x^2+1})}{\red{x^2+1}} =12ln⁡∣x2+1∣+c=\dfrac{1}{2}\ln|x^2+1|+c let x=tan⁡θ⇒dx=sec⁡2θdθlet\space x = \tan\theta\Rightarrow dx=\sec^2\theta d\theta ⇒I=∫tan⁡θsec⁡2θ1+tan⁡2θ=∫tan⁡θdθ\Rightarrow I=\int\dfrac{\tan\theta \cancel{\sec^2\theta}}{\cancel{1+\tan^2\theta}}=\int\tan\theta d\theta ⇒∫tan⁡θdθ=12ln⁡∣tan⁡2+1∣+c\Rightarrow \int\tan\theta d\theta=\dfrac{1}{2}\ln|\tan^2+1|+c =12ln⁡∣sec⁡2θ∣+c=12ln⁡∣cos⁡−2θ∣+c=\dfrac{1}{2}\ln|\sec^2\theta|+c=\dfrac{1}{2}\ln|\cos^{-2}\theta|+c =−22ln⁡∣cos⁡θ∣+c=c−ln⁡∣cos⁡θ∣=\dfrac{-2}{2}\ln|\cos\theta|+c=c-\ln|\cos\theta| ⇒∫tan⁡θdθ=c−ln⁡∣cos⁡θ∣\Rightarrow \boxed{\int\tan\theta d\theta=c-\ln|\cos\theta|}

#Calculus

Note by Zakir Husain
2 months, 1 week ago

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