1. Problem: Evaluate the integral $$\int xe^{2x} \, dx$$ using integration by parts. Formula: Integration by parts states $$\int u \, dv = uv - \int v \, du$$. Step 1: Choose $$u = x$$ (so $$du = dx$$) and $$dv = e^{2x} dx$$ (so $$v = \frac{e^{2x}}{2}$$). Step 2: Apply the formula: $$\int xe^{2x} dx = x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} dx = \frac{x e^{2x}}{2} - \frac{1}{2} \int e^{2x} dx$$. Step 3: Integrate $$\int e^{2x} dx = \frac{e^{2x}}{2}$$. Step 4: Substitute back: $$\frac{x e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + C = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C$$. --- 2. Problem: Evaluate $$\int x \cos(3x) \, dx$$. Step 1: Let $$u = x$$, $$du = dx$$; $$dv = \cos(3x) dx$$, $$v = \frac{\sin(3x)}{3}$$. Step 2: Apply integration by parts: $$x \cdot \frac{\sin(3x)}{3} - \int \frac{\sin(3x)}{3} dx = \frac{x \sin(3x)}{3} - \frac{1}{3} \int \sin(3x) dx$$. Step 3: Integrate $$\int \sin(3x) dx = -\frac{\cos(3x)}{3}$$. Step 4: Substitute: $$\frac{x \sin(3x)}{3} - \frac{1}{3} \left(-\frac{\cos(3x)}{3}\right) + C = \frac{x \sin(3x)}{3} + \frac{\cos(3x)}{9} + C$$. --- 3. Problem: Evaluate $$\int \ln x \, dx$$. Step 1: Let $$u = \ln x$$, $$du = \frac{1}{x} dx$$; $$dv = dx$$, $$v = x$$. Step 2: Apply integration by parts: $$x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int 1 dx = x \ln x - x + C$$. --- 4. Problem: Evaluate $$\int x^2 \ln x \, dx$$. Step 1: Let $$u = \ln x$$, $$du = \frac{1}{x} dx$$; $$dv = x^2 dx$$, $$v = \frac{x^3}{3}$$. Step 2: Apply integration by parts: $$\frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 dx$$. Step 3: Integrate $$\int x^2 dx = \frac{x^3}{3}$$. Step 4: Substitute: $$\frac{x^3}{3} \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C$$. --- 5. Problem: Evaluate $$\int \arcsin x \, dx$$. Step 1: Let $$u = \arcsin x$$, $$du = \frac{1}{\sqrt{1-x^2}} dx$$; $$dv = dx$$, $$v = x$$. Step 2: Apply integration by parts: $$x \arcsin x - \int x \cdot \frac{1}{\sqrt{1-x^2}} dx$$. Step 3: Use substitution for $$\int \frac{x}{\sqrt{1-x^2}} dx$$: let $$w = 1 - x^2$$, so $$dw = -2x dx$$, then $$\int \frac{x}{\sqrt{1-x^2}} dx = -\frac{1}{2} \int w^{-1/2} dw = -\frac{1}{2} \cdot 2 w^{1/2} + C = -\sqrt{1-x^2} + C$$. Step 4: Substitute back: $$x \arcsin x + \sqrt{1-x^2} + C$$. --- 6. Problem: Evaluate $$\int x^2 e^{-x} dx$$. Step 1: Let $$u = x^2$$, $$du = 2x dx$$; $$dv = e^{-x} dx$$, $$v = -e^{-x}$$. Step 2: Apply integration by parts: $$-x^2 e^{-x} + \int 2x e^{-x} dx$$. Step 3: Evaluate $$\int 2x e^{-x} dx$$ by parts again: Let $$u = 2x$$, $$du = 2 dx$$; $$dv = e^{-x} dx$$, $$v = -e^{-x}$$. Step 4: Then $$-2x e^{-x} + \int 2 e^{-x} dx = -2x e^{-x} - 2 e^{-x} + C$$. Step 5: Substitute back: $$-x^2 e^{-x} - 2x e^{-x} - 2 e^{-x} + C = -e^{-x} (x^2 + 2x + 2) + C$$. --- 7. Problem: Evaluate $$\int e^x \sin x \, dx$$. Step 1: Use integration by parts twice or use formula for $$\int e^{ax} \sin(bx) dx$$. Step 2: Let $$I = \int e^x \sin x dx$$. Step 3: Integrate by parts: Let $$u = \sin x$$, $$du = \cos x dx$$; $$dv = e^x dx$$, $$v = e^x$$. Step 4: Then $$I = e^x \sin x - \int e^x \cos x dx$$. Step 5: Let $$J = \int e^x \cos x dx$$. Step 6: Integrate $$J$$ by parts: Let $$u = \cos x$$, $$du = -\sin x dx$$; $$dv = e^x dx$$, $$v = e^x$$. Step 7: Then $$J = e^x \cos x + \int e^x \sin x dx = e^x \cos x + I$$. Step 8: Substitute back: $$I = e^x \sin x - J = e^x \sin x - (e^x \cos x + I) = e^x \sin x - e^x \cos x - I$$. Step 9: Solve for $$I$$: $$2I = e^x (\sin x - \cos x) \Rightarrow I = \frac{e^x (\sin x - \cos x)}{2} + C$$. --- 8. Problem: Evaluate $$\int x \arctan x \, dx$$. Step 1: Let $$u = \arctan x$$, $$du = \frac{1}{1+x^2} dx$$; $$dv = x dx$$, $$v = \frac{x^2}{2}$$. Step 2: Apply integration by parts: $$\frac{x^2}{2} \arctan x - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} dx = \frac{x^2}{2} \arctan x - \frac{1}{2} \int \frac{x^2}{1+x^2} dx$$. Step 3: Simplify the integrand: $$\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}$$. Step 4: Then $$\int \frac{x^2}{1+x^2} dx = \int 1 dx - \int \frac{1}{1+x^2} dx = x - \arctan x + C$$. Step 5: Substitute back: $$\frac{x^2}{2} \arctan x - \frac{1}{2} (x - \arctan x) + C = \frac{x^2 + 1}{2} \arctan x - \frac{x}{2} + C$$. --- 9. Problem: Evaluate $$\int \ln(x^2 + 1) dx$$. Step 1: Let $$u = \ln(x^2 + 1)$$, $$du = \frac{2x}{x^2 + 1} dx$$; $$dv = dx$$, $$v = x$$. Step 2: Apply integration by parts: $$x \ln(x^2 + 1) - \int x \cdot \frac{2x}{x^2 + 1} dx = x \ln(x^2 + 1) - 2 \int \frac{x^2}{x^2 + 1} dx$$. Step 3: Simplify the integrand: $$\frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$$. Step 4: Then $$\int \frac{x^2}{x^2 + 1} dx = \int 1 dx - \int \frac{1}{x^2 + 1} dx = x - \arctan x + C$$. Step 5: Substitute back: $$x \ln(x^2 + 1) - 2(x - \arctan x) + C = x \ln(x^2 + 1) - 2x + 2 \arctan x + C$$. --- 10. Problem: Evaluate $$\int x^3 e^{x^2} dx$$. Step 1: Use substitution: let $$w = x^2$$, so $$dw = 2x dx$$, then $$x^3 dx = x^2 \cdot x dx = w \cdot \frac{dw}{2} = \frac{w dw}{2}$$. Step 2: Rewrite integral: $$\int x^3 e^{x^2} dx = \int w e^w \frac{dw}{2} = \frac{1}{2} \int w e^w dw$$. Step 3: Use integration by parts on $$\int w e^w dw$$: Let $$u = w$$, $$du = dw$$; $$dv = e^w dw$$, $$v = e^w$$. Step 4: Then $$w e^w - \int e^w dw = w e^w - e^w + C = e^w (w - 1) + C$$. Step 5: Substitute back: $$\frac{1}{2} e^w (w - 1) + C = \frac{1}{2} e^{x^2} (x^2 - 1) + C$$. --- 11. Problem: Evaluate $$\int \sin^3 x \cos^2 x \, dx$$. Step 1: Use $$\sin^3 x = \sin x (1 - \cos^2 x)$$. Step 2: Substitute: $$\int \sin x (1 - \cos^2 x) \cos^2 x dx = \int \sin x (\cos^2 x - \cos^4 x) dx$$. Step 3: Let $$u = \cos x$$, $$du = -\sin x dx$$, so $$-du = \sin x dx$$. Step 4: Rewrite integral: $$- \int (u^2 - u^4) du = - \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C = - \frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$$. --- 12. Problem: Evaluate $$\int \cos^4 x \, dx$$. Step 1: Use power-reduction formula: $$\cos^4 x = \left( \cos^2 x \right)^2 = \left( \frac{1 + \cos 2x}{2} \right)^2 = \frac{1}{4} (1 + 2 \cos 2x + \cos^2 2x)$$. Step 2: Use power-reduction again on $$\cos^2 2x$$: $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$. Step 3: Substitute: $$\int \cos^4 x dx = \frac{1}{4} \int \left(1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) dx = \frac{1}{4} \int \left( \frac{3}{2} + 2 \cos 2x + \frac{\cos 4x}{2} \right) dx$$. Step 4: Integrate term by term: $$\frac{1}{4} \left( \frac{3}{2} x + \sin 2x + \frac{\sin 4x}{8} \right) + C = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$$. --- 13. Problem: Evaluate $$\int \sin^2 x \cos^2 x \, dx$$. Step 1: Use identity: $$\sin^2 x \cos^2 x = \left( \sin x \cos x \right)^2 = \left( \frac{\sin 2x}{2} \right)^2 = \frac{\sin^2 2x}{4}$$. Step 2: Use power-reduction: $$\sin^2 2x = \frac{1 - \cos 4x}{2}$$. Step 3: Substitute: $$\int \sin^2 x \cos^2 x dx = \frac{1}{4} \int \frac{1 - \cos 4x}{2} dx = \frac{1}{8} \int (1 - \cos 4x) dx$$. Step 4: Integrate: $$\frac{1}{8} (x - \frac{\sin 4x}{4}) + C = \frac{x}{8} - \frac{\sin 4x}{32} + C$$. --- 14. Problem: Evaluate $$\int \tan^3 x \sec x \, dx$$. Step 1: Rewrite $$\tan^3 x = \tan^2 x \tan x = (\sec^2 x - 1) \tan x$$. Step 2: Substitute: $$\int (\sec^2 x - 1) \tan x \sec x dx$$. Step 3: Use substitution $$u = \sec x$$, $$du = \sec x \tan x dx$$. Step 4: Then integral becomes: $$\int (u^2 - 1) du = \frac{u^3}{3} - u + C = \frac{\sec^3 x}{3} - \sec x + C$$. --- 15. Problem: Evaluate $$\int \sec^4 x \, dx$$. Step 1: Use identity: $$\sec^4 x = (\sec^2 x)^2 = (1 + \tan^2 x)^2 = 1 + 2 \tan^2 x + \tan^4 x$$. Step 2: Rewrite integral: $$\int (1 + 2 \tan^2 x + \tan^4 x) dx$$. Step 3: Use substitution $$t = \tan x$$, $$dt = \sec^2 x dx$$. Step 4: Express $$dx$$: $$dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2}$$. Step 5: Integral becomes complicated; better to use reduction formula or rewrite: Use formula: $$\int \sec^n x dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x dx$$ for $$n=4$$. Step 6: Then $$\int \sec^4 x dx = \frac{\sec^2 x \tan x}{3} + \frac{2}{3} \int \sec^2 x dx = \frac{\sec^2 x \tan x}{3} + \frac{2}{3} \tan x + C$$. --- 16. Problem: Evaluate $$\int \frac{\sin^3 x}{\cos x} dx$$. Step 1: Rewrite $$\sin^3 x = \sin x (1 - \cos^2 x)$$. Step 2: Substitute: $$\int \frac{\sin x (1 - \cos^2 x)}{\cos x} dx = \int \sin x \left( \frac{1}{\cos x} - \cos x \right) dx = \int \sin x \sec x dx - \int \sin x \cos x dx$$. Step 3: Simplify: $$\int \tan x dx - \int \sin x \cos x dx$$. Step 4: Integrate: $$\int \tan x dx = -\ln |\cos x| + C$$. Step 5: For $$\int \sin x \cos x dx$$, use substitution $$u = \sin x$$, $$du = \cos x dx$$: $$\int u du = \frac{u^2}{2} + C = \frac{\sin^2 x}{2} + C$$. Step 6: Final answer: $$-\ln |\cos x| - \frac{\sin^2 x}{2} + C$$. --- 17. Problem: Evaluate $$\int \sin(2x) \cos(3x) dx$$. Step 1: Use product-to-sum formula: $$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$$. Step 2: Substitute: $$\int \sin(2x) \cos(3x) dx = \frac{1}{2} \int [\sin(5x) + \sin(-x)] dx = \frac{1}{2} \int [\sin(5x) - \sin x] dx$$. Step 3: Integrate: $$\frac{1}{2} \left(-\frac{\cos 5x}{5} + \cos x \right) + C = -\frac{\cos 5x}{10} + \frac{\cos x}{2} + C$$. --- 18. Problem: Evaluate $$\int \tan^2 x \sec^2 x dx$$. Step 1: Use substitution $$u = \tan x$$, $$du = \sec^2 x dx$$. Step 2: Integral becomes: $$\int u^2 du = \frac{u^3}{3} + C = \frac{\tan^3 x}{3} + C$$. --- 19. Problem: Evaluate $$\int \frac{dx}{\sin x \cos^2 x}$$. Step 1: Rewrite integrand: $$\frac{1}{\sin x \cos^2 x} = \frac{\csc x}{\cos^2 x} = \csc x \sec^2 x$$. Step 2: Use substitution $$t = \tan x$$, $$dt = \sec^2 x dx$$. Step 3: Then $$\int \csc x \sec^2 x dx = \int \csc x dt$$, but $$\csc x$$ is in terms of $$x$$, so express $$\csc x$$ in terms of $$t$$: Since $$t = \tan x = \frac{\sin x}{\cos x}$$, then $$\sin x = \frac{t}{\sqrt{1+t^2}}$$, so $$\csc x = \frac{1}{\sin x} = \frac{\sqrt{1+t^2}}{t}$$. Step 4: Integral becomes: $$\int \frac{\sqrt{1+t^2}}{t} dt$$. Step 5: This integral is non-elementary in simple form; alternatively, use substitution or rewrite original integral differently. Step 6: Alternatively, rewrite original integral as: $$\int \frac{dx}{\sin x \cos^2 x} = \int \frac{\sec^2 x}{\sin x} dx$$. Step 7: Use substitution $$u = \tan x$$, $$du = \sec^2 x dx$$, so $$\int \frac{\sec^2 x}{\sin x} dx = \int \frac{1}{\sin x} \sec^2 x dx = \int \frac{1}{\sin x} du$$. Step 8: Express $$\sin x$$ in terms of $$u$$: $$\sin x = \frac{u}{\sqrt{1+u^2}}$$. Step 9: Then $$\int \frac{1}{\sin x} du = \int \frac{\sqrt{1+u^2}}{u} du$$, same as step 4. Step 10: Use substitution $$u = \sinh t$$, then $$\sqrt{1+u^2} = \cosh t$$, and $$du = \cosh t dt$$. Step 11: Integral becomes: $$\int \frac{\cosh t}{\sinh t} \cosh t dt = \int \frac{\cosh^2 t}{\sinh t} dt$$. Step 12: Use identity $$\cosh^2 t = 1 + \sinh^2 t$$: $$\int \frac{1 + \sinh^2 t}{\sinh t} dt = \int \left( \frac{1}{\sinh t} + \sinh t \right) dt$$. Step 13: Integrate: $$\int \frac{1}{\sinh t} dt + \int \sinh t dt = \int \csch t dt + \cosh t + C$$. Step 14: Integral of $$\csch t$$ is $$\ln \left| \tanh \frac{t}{2} \right| + C$$. Step 15: Substitute back to $$x$$: $$t = \sinh^{-1} u = \sinh^{-1} (\tan x)$$. Step 16: Final answer: $$\ln \left| \tanh \frac{\sinh^{-1} (\tan x)}{2} \right| + \cosh (\sinh^{-1} (\tan x)) + C$$. --- 20. Problem: Evaluate $$\int \sin^4 x \, dx$$. Step 1: Use power-reduction: $$\sin^4 x = (\sin^2 x)^2 = \left( \frac{1 - \cos 2x}{2} \right)^2 = \frac{1}{4} (1 - 2 \cos 2x + \cos^2 2x)$$. Step 2: Use power-reduction on $$\cos^2 2x$$: $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$. Step 3: Substitute: $$\int \sin^4 x dx = \frac{1}{4} \int \left(1 - 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) dx = \frac{1}{4} \int \left( \frac{3}{2} - 2 \cos 2x + \frac{\cos 4x}{2} \right) dx$$. Step 4: Integrate term by term: $$\frac{1}{4} \left( \frac{3}{2} x - \sin 2x + \frac{\sin 4x}{8} \right) + C = \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$$. --- q_count: 20