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Evaluate \(\int \frac{\cos (x)}{7+5 \sin (x)-\cos ^{2}(x)} d x\)

\(u=\sin (x) \longrightarrow d u=\cos (x) d x\)

\(\cos ^{2}(x)=\left(1-\sin ^{2}(x)\right)\)

\(\cos ^{2} x+\sin ^{2} x=1\)

\(I=\int \frac{\cos (x)}{7+5 \sin (x)-\left(1-\sin ^{2}(x)\right)}=\int \frac{d u}{7+5 u-\left(1-u^{2}\right)}\)

\(=\int \frac {d u}{7+5 u -1+u^{2}}=\int \frac{d u}{u^{2}+5 u+6}\)

\(I=\int \frac{d u}{u^{2}+5 u+6}=\int \frac{d u}{(u+2)(u+3)}=\int \frac{A}{u+2} d u+\int \frac{B}{u+3} d u\)

\(A=(u+2) \cdot\left.fu \right|_{u=-2}=(u+2) \cdot\left.\frac{1}{(u+2)(u+3)}\right|_{u=-2}=\left.\frac{1}{u+3}\right|_{u=-2}=1\)

\(B=(u+3) f\left.(u)\right|_{u=-3}=(u+3) \cdot\left.\frac{1}{(u+2) (u+3)}\right|_{u=-3}=-1\)

\(I=\int \frac{du}{u^{2}+{5u}+6}=\int \frac{1}{u+2} d u+\int \frac{-1}{u+3} d u\)

\(=\ln |u+2|-\ln |u+3|+c\)

\(\int \frac{\cos (x)}{7+5 \sin (x)-\cos ^{2}(x)} d x=\ln |\sin (x)+2|-\ln |\sin (x)+3|+c\)

\(\ln a-\ln b=\ln \left|\frac{a}{b}\right|\)

\(\int \frac{\cos (x)}{7+5 \sin (x)-\cos ^{2}(x)}dx=\ln \left|\frac{\sin (x)+2}{\sin (x)+3}\right|+c\)

Evaluate \(I=\int \frac{\sqrt{x}}{1+\sqrt[3]{x}} d x\)

\(\sqrt[2]{x}, \sqrt[3]{x}=6\)

let \(t^{6}=x \Rightarrow d x=6 t^{5} d t\)

\(\sqrt{x}=\sqrt{t^{6}}=t^{\frac{6}{2}}=t^{3} \quad \sqrt[3]{x}=\sqrt[3]{t^{6}}=t^{\frac{6}{3}}=t^{2}\)

\(I=\int \frac{\sqrt{x}}{1+\sqrt[3]{x}} d x=\int \frac{t^{3}}{1+t^{2}}\left(6 t^{5} d t\right)\)

\(=6 \int \frac{t^{8}}{1+t^{2}} d t\)

long division  قسمة مطولة

\(I=\int \frac{\sqrt{x}}{1+\sqrt[3]{3 x}} d x=6 \int \frac{t^{8}}{1+t^{2}} d t=6 \int\left(t^{6}-t^{4}+t^{2}+\frac{1}{t^{2}+1}\right) d t\)

\(\int \frac{1}{x^{2}+1}=\tan ^{-1}(x)\)

\(I=6\left[\frac{t^{7}}{7}-\frac{t^{5}}{5}+\frac{t^{3}}{3}-t+\tan ^{-1}(t)\right]+c\)

\(t^{6 / 6}=x^{1 / 6}\)

\(t=x^{\frac{1}{6}}\)

\(I=6\left[\frac{{x^{\frac{1}{6}\cdot {7}}}}{7}-\frac{x^{\frac {1}{6}\cdot 5}}{5}+\frac{x^{\frac{1}{6}\cdot 3}}{3}-x^\frac{1}{6}+\tan ^{-1}\left(x^{\frac{1}{6}}\right)+c\right]\)

\(I=\frac{6}{7} x^{\frac{7}{6}}-\frac{6}{5} x^{\frac{5}{6}}+\frac{6}{3} x^{\frac{1}{2}}-6x^{\frac{1}{6}}+6\tan ^{-1}\left(x^{\frac{1}{6}}\right)+c\)

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