JEE Main 2019 — Mathematics Calculus

Given integral, $\displaystyle I={\int }_{0}^{1}x{\mathrm{cot}}^{-1}(1-{x}^{2}+{x}^{4})dx$I=∫01​xcot−1(1−x2+x4)dx

Put ${x}^{2}=t, \Rightarrow 2xdx=dt$x2=t,⇒2xdx=dt

$$\Rightarrow I=\frac{1}{2}{\int }_{0}^{1}{\mathrm{cot}}^{-1}(1-t+{t}^{2})dt$$⇒I=21​∫01​cot−1(1−t+t2)dt

$$\Rightarrow I=\frac{1}{2}{\int }_{0}^{1}{\mathrm{tan}}^{-1}(\frac{1}{1-t+{t}^{2}})dt$$⇒I=21​∫01​tan−1(1−t+t21​)dt

$$\Rightarrow I=\frac{1}{2}{\int }_{0}^{1}{\mathrm{tan}}^{-1}(\frac{t-(t-1)}{1+t(t-1)})dt$$⇒I=21​∫01​tan−1(1+t(t−1)t−(t−1)​)dt

Using ${\mathrm{tan}}^{-1}a-{\mathrm{tan}}^{-1}b={\mathrm{tan}}^{-1}(\frac{a-b}{1+a\cdot b}),$tan−1a−tan−1b=tan−1(1+a⋅ba−b​), we get

$$\Rightarrow I=\frac{1}{2}[{\int }_{0}^{1}({\mathrm{tan}}^{-1}t)dt-{\int }_{0}^{1}{\mathrm{tan}}^{-1}(t-1)dt]$$⇒I=21​[∫01​(tan−1t)dt−∫01​tan−1(t−1)dt]

Using $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx,$∫ab​f(x)dx=∫ab​f(a+b−x)dx, we get

$$\Rightarrow I=\frac{1}{2}[{\int }_{0}^{1}({\mathrm{tan}}^{-1}t)dt-{\int }_{0}^{1}{\mathrm{tan}}^{-1}((1-t)-1)dt]$$⇒I=21​[∫01​(tan−1t)dt−∫01​tan−1((1−t)−1)dt]

$$\Rightarrow I=\frac{1}{2}[{\int }_{0}^{1}({\mathrm{tan}}^{-1}t)dt-{\int }_{0}^{1}{\mathrm{tan}}^{-1}(-t)dt]$$⇒I=21​[∫01​(tan−1t)dt−∫01​tan−1(−t)dt]

Now, using ${\mathrm{tan}}^{-1}(-a)=-{\mathrm{tan}}^{-1}a,$tan−1(−a)=−tan−1a, we get

$$\Rightarrow I=\frac{1}{2}[{\int }_{0}^{1}({\mathrm{tan}}^{-1}t)dt+{\int }_{0}^{1}({\mathrm{tan}}^{-1}t)dt]$$⇒I=21​[∫01​(tan−1t)dt+∫01​(tan−1t)dt]

$$\Rightarrow I=\frac{1}{2}[2{\int }_{0}^{1}{tan}^{-1}tdt]$$⇒I=21​[2∫01​tan−1tdt]

$$\Rightarrow I={\int }_{0}^{1}(1\cdot {\mathrm{tan}}^{-1}t)dt$$⇒I=∫01​(1⋅tan−1t)dt

Now applying the integration by parts i.e. $\displaystyle {\int }_{a}^{b}(u\cdot v)dx=u{\int }_{a}^{b}vdx-{\int }_{a}^{b}[\frac{d}{dx}(u)\int vdx]dx+c$∫ab​(u⋅v)dx=u∫ab​vdx−∫ab​[dxd​(u)∫vdx]dx+c

$$\Rightarrow I={{tan}^{-1}t\cdot t|}_{0}^{1}-{\int }_{0}^{1}(\frac{1}{1+{t}^{2}}\cdot t)dt$$⇒I=tan−1t⋅t∣01​−∫01​(1+t21​⋅t)dt

Put $1+{t}^{2}=u,\Rightarrow 2tdt=du$1+t2=u,⇒2tdt=du

$$\Rightarrow I={\mathrm{tan}}^{-1}(1)\cdot 1-0-\frac{1}{2}{\int }_{1}^{2}(\frac{1}{u})du$$⇒I=tan−1(1)⋅1−0−21​∫12​(u1​)du

$$\Rightarrow I=\frac{\pi }{4}-\frac{1}{2}{\mathrm{log}}_{2}(u){​|}_{1}^{2}$$⇒I=4π​−21​log2​(u)​∣12​

$$\Rightarrow I=\frac{\pi }{4}-\frac{1}{2}{\mathrm{log}}_{e}2.$$⇒I=4π​−21​loge​2.