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Question Number 118733 by mohammad17 last updated on 19/Oct/20

Commented by bemath last updated on 19/Oct/20

$$\left(4\right)\frac{d}{dx}[\int {}_{x^3}^{2x^2}\cos (2t^3+1)dt]=$$$$4x\cos (2.\left(8x^6\right)+1)-3x^2\cos (2.x^9+1)$$$$=4x\cos (16x^6+1)-3x^2\cos (2x^9+1)$$

Answered by TANMAY PANACEA last updated on 19/Oct/20

$$5)0<{\int }_0^{\mathrm{\infty }}\frac{sinx}{1+x^2}<{\int }_0^{\mathrm{\infty }}\frac{1}{1+x^2}$$$$0<I<\mid {tan}^{-1}x{\mid }_0^{\mathrm{\infty }}$$$$0<I<\frac{\pi }{2}$$$$$$

Answered by TANMAY PANACEA last updated on 19/Oct/20

$$4)I\left(x\right)={\int }_{x^3}^{2x^2}cos(2t^3+1)dt$$$$\frac{dI}{dx}={\int }_{x^3}^{2x^2}\frac{\partial }{\partial x}cos(2t^3+1)dx+cos\{{\left(2x^2\right)}^3+1\}\frac{d\left(2x^2\right)}{dx}-cos\{2{\left(x^3\right)}^3+1\}\frac{d\left(x^3\right)}{dx}$$$$=0+4xcos(2x^6+1)-3x^{2}cos(2x^9+1)$$$$★wait...$$

Answered by TANMAY PANACEA last updated on 19/Oct/20

$$i)I={\int }_0^{\frac{\pi }{2}}\frac{{sin}^{3}x}{{cos}^{3}x+{sin}^{3}x}dx$$$$I={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{3}x}{{sin}^{3}x+{cos}^{3}x}dx[using{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx]$$$$2I={\int }_0^{\frac{\pi }{2}}dx\to I=\frac{\pi }{4}$$

Answered by TANMAY PANACEA last updated on 19/Oct/20

$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+cotx}={\int }_0^{\frac{\pi }{2}}\frac{sinx}{cosx+sinx}dx$$$$I={\int }_0^{\frac{\pi }{2}}\frac{cosx}{sinx+cosx}dx$$$$2I={\int }_0^{\frac{\pi }{2}}dx$$$$I=\frac{\pi }{4}$$

Commented by mnjuly1970 last updated on 19/Oct/20

$$thankyoumrtanmay..$$$$.m.n.july.1970..$$

Commented by TANMAY PANACEA last updated on 19/Oct/20

$$mostwelcomesir$$

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