Integration Questions

Solution : We have, I = $\int$ $sin^{-1}(cos x)$ dx By using integration formula, cos x = $sin({\pi\over 2} – x)$ I = $\int$ $sin^{-1}[sin({\pi\over 2} – x)]$ dx I = $\int$ $({\pi\over 2} – x)$ dx I = ${\pi\over 2}x$ – $x^2\over 2$ + C Similar Questions What is the integration of sin inverse […]

Solution : We have, I = $sin^{-1}\sqrt{x}$ . 1 dx By Applying integration by parts, Taking $sin^{-1}\sqrt{x}$ as first function and 1 as second function. Then I = $sin^{-1}\sqrt{x}$ $\int$ 1 dx – $\int$ {$d\over dx$$sin^{-1}\sqrt{x}$ $\int$ 1 dx } dx I = x$sin^{-1}\sqrt{x}$ – $\int$ $1\over 2\sqrt{(1-x)}\sqrt{x}$ . x dx I = x$sin^{-1}\sqrt{x}$ –

Solution : We have, I = $(sin^{-1}x)^2$ dx Let $sin^{-1}x$ = t, Then, x = sin t $\implies$ dx = cos t dt $\therefore$ I = $\int$ $(sin^{-1}x)^2$ dx I = $\int$ $t^2$ cos t dt Applying integration by parts and, Taking $t^2$ as first function and cos t as second function, I = $t^2$

Solution : We have, I = $\int$  $x sin^{-1} x$ dx By using integration by parts formula, I = $sin^{-1} x$ $x^2\over 2$ – $\int$ $1\over \sqrt{1 – x^2}$ $\times$ $x^2\over 2$ dx I =  $x^2\over 2$ $sin^{-1} x$ + $1\over 2$ $\int$ $-x^2\over \sqrt{1 – x^2}$ dx = $x^2\over 2$ $sin^{-1} x$ + \(1\over

Solution : Let I = $\int$ $tan^{-1}\sqrt{x}$.1 dx By Applying integration by parts, Taking $tan^{-1}\sqrt{x}$ as first function and 1 as second function. Then I = $tan^{-1}\sqrt{x}$ $\int$ 1 dx – $\int$ {$d\over dx$$tan^{-1}\sqrt{x}$ $\int$ 1 dx } dx I = x$tan^{-1}\sqrt{x}$ – $\int$ $1\over 2(1+x)\sqrt{x}$ . x dx Let $\sqrt{x}$ = t $1\over 2\sqrt{x}$

Solution : Let I = $\int$ x$tan^{-1}x$ dx By using Integration by parts rule, Taking tan inverse x as first function and x as second function. Then, I = ($tan^{-1}x$) $\int$ x dx – $\int${${d\over dx}$($tan^{-1}x$ $\int$ x dx} dx I = ($tan^{-1}x$)$x^2\over 2$ – $\int$${1\over 1 + x^2}$ $\times$ $x^2\over 2$ dx $\implies$ I

Solution : Let I = $\int_{0}^{\pi/2}$ log(sinx)dx    …….(i) then I = $\int_{0}^{\pi/2}$ $log(sin({\pi\over 2}-x))$dx = $\int_{0}^{\pi/2}$ log(cosx)dx     …….(ii) Adding (i) and (ii), we get 2I = $\int_{0}^{\pi/2}$ log(sinx)dx + $\int_{0}^{\pi/2}$ log(cosx)dx = $\int_{0}^{\pi/2}$ (log(sinx)dx + log(cosx)) $\implies$ $\int_{0}^{\pi/2}$ log(sinxcosx)dx = $\int_{0}^{\pi/2}$ $log({2sinxcosx\over 2})$dx = $\int_{0}^{\pi/2}$ $log({sin2x\over 2})$dx = $\int_{0}^{\pi/2}$ log(sin2x)dx – $\int_{0}^{\pi/2}$ log(2)dx

Solution : I = $\int$ $cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}$ = $\int$ $cos^4xdx\over {sin^6x{(1 + cot^5x)^{3\over 5}}}$ = $\int$ $cot^4xcosec^2xdx\over {{(1 + cot^5x)^{3\over 5}}}$ Put $1+cot^5x$ = t $5cot^4xcosec^2x$dx = -dt = -$1\over 5$ $\int$ $dt\over {t^{3/5}}$ = -$1\over 2$ $t^{2/5}$ + C = -$1\over 2$ ${(1+cot^5x)}^{2/5}$ + C Similar Questions What is the integration

Solution : I = $\int$ $dx\over {3sinx + 4cosx}$ = $\int$ $dx\over {3[{2tan{x\over 2}\over {1+tan^2{x\over 2}}}] + 4[{1-tan^2{x\over 2}\over {1+tan^2{x\over 2}}}]}$ = $\int$ $sec^2{x\over 2}dx\over {4+6tan{x\over 2}-4tan^2{x\over 2}}$ let $tan{x\over 2}$ = t, $\therefore$  ${1\over 2}sec^2{x\over 2}$dx = dt so I = $\int$ $2dt\over {4+6t-4t^2}$ = $1\over 2$ $\int$ $dt\over {1-(t^2-{3\over 2}t})$ = $1\over 2$