Question-118733

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

Commented by bemath last updated on 19/Oct/20

(4) (d/dx) [ ∫ _x^3 ^(2x^2 ) cos (2t^3 +1) dt ] = 4x cos (2.(8x^6 )+1)−3x^2 cos (2.x^9 +1) =4x cos (16x^6 +1)−3x^2 cos (2x^9 +1)

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

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

5)0<∫_0 ^∞ ((sinx)/(1+x^2 ))<∫_0 ^∞ (1/(1+x^2 )) 0<I<∣tan^(−1) x∣_0 ^∞ 0<I<(π/2)

$$\left.\mathrm{5}\right)\mathrm{0}<\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} }<\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{0}<{I}<\mid{tan}^{−\mathrm{1}} {x}\mid_{\mathrm{0}} ^{\infty} \\ $$$$\mathrm{0}<{I}<\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$

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

4)I(x) =∫_x^3 ^(2x^2 ) cos(2t^3 +1)dt (dI/dx)=∫_x^3 ^(2x^2 ) (∂/∂x)cos(2t^3 +1)dx+cos{(2x^2 )^3 +1}((d(2x^2 ))/dx)−cos{2(x^3 )^3 +1}((d(x^3 ))/dx) =0+4xcos(2x^6 +1)−3x^2 cos(2x^9 +1) ★wait...

$$\left.\mathrm{4}\right){I}\left({x}\right)\:\:\:=\int_{{x}^{\mathrm{3}} } ^{\mathrm{2}{x}^{\mathrm{2}} } {cos}\left(\mathrm{2}{t}^{\mathrm{3}} +\mathrm{1}\right){dt} \\ $$$$\frac{{dI}}{{dx}}=\int_{{x}^{\mathrm{3}} } ^{\mathrm{2}{x}^{\mathrm{2}} } \:\frac{\partial}{\partial{x}}{cos}\left(\mathrm{2}{t}^{\mathrm{3}} +\mathrm{1}\right){dx}+{cos}\left\{\left(\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{1}\right\}\frac{{d}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{dx}}−{cos}\left\{\mathrm{2}\left({x}^{\mathrm{3}} \right)^{\mathrm{3}} +\mathrm{1}\right\}\frac{{d}\left({x}^{\mathrm{3}} \right)}{{dx}} \\ $$$$=\mathrm{0}+\mathrm{4}{xcos}\left(\mathrm{2}{x}^{\mathrm{6}} +\mathrm{1}\right)−\mathrm{3}{x}^{\mathrm{2}} {cos}\left(\mathrm{2}{x}^{\mathrm{9}} +\mathrm{1}\right) \\ $$$$\bigstar{wait}… \\ $$

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

i)I=∫_0 ^(π/2) ((sin^3 x)/(cos^3 x+sin^3 x))dx I=∫_0 ^(π/2) ((cos^3 x)/(sin^3 x+cos^3 x))dx [using ∫_0 ^a f(x)dx =∫_0 ^a f(a−x)dx] 2I=∫_0 ^(π/2) dx→I=(π/4)

$$\left.{i}\right){I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{3}} {x}}{{cos}^{\mathrm{3}} {x}+{sin}^{\mathrm{3}} {x}}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{3}} {x}}{{sin}^{\mathrm{3}} {x}+{cos}^{\mathrm{3}} {x}}{dx}\:\:\left[{using}\:\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx}\right] \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\rightarrow{I}=\frac{\pi}{\mathrm{4}} \\ $$

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

I=∫_0 ^(π/2) (dx/(1+cotx))=∫_0 ^(π/2) ((sinx)/(cosx+sinx))dx I=∫_0 ^(π/2) ((cosx)/(sinx+cosx))dx 2I=∫_0 ^(π/2) dx I=(π/4)

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{cotx}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinx}}{{cosx}+{sinx}}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cosx}}{{sinx}+{cosx}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$${I}=\frac{\pi}{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 19/Oct/20

$${thank}\:{you}\:{mr}\:{tanmay}.. \\ $$$$.{m}.{n}.{july}.\mathrm{1970}.. \\ $$

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

$${most}\:{welcome}\:{sir} \\ $$

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