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Question Number 32363 by prof Abdo imad last updated on 23/Mar/18

$$letconsiderthefunction$$$$f(x,\theta )={\int }_x^{x^2}ln(2+sin\theta cost)dt$$$$calculate\frac{\partial f}{\partial x}(x,\theta )and\frac{\partial f}{\partial \theta }(x,\theta ).$$

Commented by prof Abdo imad last updated on 25/Mar/18

$$\frac{\partial f}{\partial x}(x,\theta )=2xln(2+sin\theta cos\left(x^2\right))-ln(2+sin\theta cosx)$$$$\frac{\partial f}{\partial \theta }(x,\theta )={\int }_x^{x^2}\frac{costcos\theta }{2+costsin\theta }dtch.tan\left(\frac{t}{2}\right)=ugive$$$$\frac{\partial f}{\partial \theta }={\int }_{tan\left(\frac{x}{2}\right)}^{tan\left(\frac{x^2}{2}\right)}\frac{cos\theta \frac{1-u^2}{1+u^2}}{2+sin\theta \frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$$$$={\int }_{tan\left(\frac{x}{2}\right)}^{tan\left(\frac{x^2}{2}\right)}\frac{(1-u^2)cos\theta }{(1+u^2)(2+(1-u^2)sin\theta )}du$$$$={\int }_{tan\left(\frac{x}{2}\right)}^{tan\left(\frac{x^2}{2}\right)}\frac{\alpha -\alpha u^2}{(1+u^2)(2+\beta -\beta u^2)}duwhit\alpha =cos\theta$$$$and\beta =sin\theta$$$$={\int }_{tan\left(\frac{x}{2}\right)}^{tan\left(\frac{x^2}{2}\right)}\frac{-\alpha u^2+\alpha }{(1+u^2)(-\beta u^2+\beta +2)}duletdecompose$$$$f\left(u\right)=\frac{\alpha u^2-\alpha }{(1+u^2)(\beta u^2-\beta -2)}$$$$=\frac{\alpha u^2-\alpha }{\beta (1+u^2)(u^2-{\left(\sqrt{\frac{\beta +2}{\beta }}\right)}^2)}$$$$=\frac{a}{(u-\sqrt{\frac{\beta +2}{\beta }})}+\frac{b}{u+\sqrt{\frac{\beta +2}{\beta }}}+\frac{cu+d}{u^2+1}...be$$$$conyinued....$$

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