1. (s-10) By using substitution solve the following lodefinite integrate (a) (b) int (x^3dx)/(sqrt [3](2+x^4)) (c) int xcosx^2dx int (dx)/(sqrt (x)(1-sqrt (x))^3) 2. (s-10) By using substitution solve the following definite integrals. (a) int _(0)^1(xdx)/(sqrt [3](1+x^2)) (b) int _(0)^(pi )/(2)(cosxdx)/(sqrt (2+sinx)) 3. (S-15) By using definite integral find area of the following region enclosed by the parabola y_(1)=x^2+2x+2 and the line y_(2)=4x+5 4. (S-15) The region bounded by the curve y_(1)=x^2-2x and the line y_(2)=2x+5 is revolved about the r-axis to generate a solid. By using definite integral (Washer Method) find the volume of the solid. 5. (S-15) Find the length of the following plane curve defined by: x=sin^2t y=cos^2t,0leqslant tleqslant (pi )/(4) 6. By using integration by parts evaluate the integrals: int x^(1)/(3)lnxdx,int x^2cosxdx 7. (S-15) By using integration by part and functional equation evaluate the integrals int e^2xcos3xdx (S-10) By using integration of rational functions evalute the integral int (x^3-2x^2+x+3)/(x^2)-4x+3dx

1. (a) $\int \frac {x^{3}dx}{\sqrt [3]{2+x^{4}}}$<br /><br />Let $u = 2 + x^4$, then $du = 4x^3 dx$, and $dx = \frac{du}{4x^3}$.<br /><br />Substitute $u$ and $dx$ into the integral:<br /><br />$\int \frac{x^3 dx}{\sqrt[3]{2+x^4}} = \int \frac{x^3 \cdot \frac{du}{4x^3}}{\sqrt[3]{u}} = \frac{1}{4} \int \frac{du}{u^{1/3}} = \frac{1}{4} \int u^{-1/3} du = \frac{1}{4} \cdot \frac{3}{2} u^{2/3} + C = \frac{3}{8} (2 + x^4)^{2/3} + C$<br /><br />(b) $\int x \cos x^2 dx$<br /><br />Let $u = x^2$, then $du = 2x dx$, and $dx = \frac{du}{2x}$.<br /><br />Substitute $u$ and $dx$ into the integral:<br /><br />$\int x \cos x^2 dx = \int x \cos u \cdot \frac{du}{2x} = \frac{1}{2} \int \cos u du = \frac{1}{2} \sin u + C = \frac{1}{2} \sin x^2 + C$<br /><br />(c) $\int \frac{dx}{\sqrt{x}(1-\sqrt{x})^3}$<br /><br />Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}} dx$, and $dx = 2u du$.<br /><br />Substitute $u$ and $dx$ into the integral:<br /><br />$\int \frac{dx}{\sqrt{x}(1-\sqrt{x})^3} = \int \frac{2u du}{u(1-u)^3} = 2 \int \frac{du}{(1-u)^3} = 2 \int (1-u)^{-3} du = 2 \cdot \frac{(1-u)^{-2}}{-2} + C = - (1-u)^{-2} + C = - (1-\sqrt{x})^{-2} + C$<br /><br />2. (a) $\int_0^1 \frac{xdx}{\sqrt[3]{1+x^2}}$<br /><br />Let $u = 1 + x^2$, then $du = 2x dx$, and $dx = \frac{du}{2x}$.<br /><br />Substitute $u$ and $dx$ into the integral:<br /><br />$\int_0^1 \frac{xdx}{\sqrt[3]{1+x^2}} = \int_0^1 \frac{x \cdot \frac{du}{2x}}{u^{1/3}} = \frac{1}{2} \int_0^1 u^{-1/3} du = \frac{1}{2} \cdot \frac{3}{2} u^{2/3} \Big|_0^1 = \frac{3}{4} \cdot 1^{2/3} - \frac{3}{4} \cdot 0^{2/3} = \frac{3}{4}$<br /><br />(b) $\int_0^{\frac{\pi}{2}} \frac{\cos x dx}{\sqrt{2+\sin x}}$<br /><br />Let $u = \sin x$, then $du = \cos x dx$, and $dx = \frac{du}{\cos x}$.<br /><br />Substitute $u$ and $dx$ into the integral:<br /><br />$\int_0^{\frac{\pi}{2}} \frac{\cos x dx}{\sqrt{2+\sin x}} = \int_0^{\frac{\pi}{2}} \frac{\cos x \cdot \frac{du}{\cos x}}{\sqrt{2+u}} = \int_0^{\frac{\pi}{2}} \frac{du}{\sqrt{2+u}} = \sqrt{2+u} \Big|_0^{\frac{\pi}{2}} = \sqrt{2+\sin x} \Big|_0^{\frac{\pi}{2}} = \sqrt{2+\sin \frac{\pi}{2}} - \sqrt{2+\sin 0} = \sqrt{3} - \sqrt{2}$<br /><br />3. The area enclosed by the parabola $y_1 = x^2 + 2x + 2$ and the line $y_2 = 4x + 5$ is given by the integral:<br /><br />$\int_{-\in