Indefinite Integration flash cards

1. 1.Evaluate the standard integral $\int \frac{dx}{a^2+x^2}$.

   $\dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)+C$

   Hint: Think inverse tangent form.

2. 2.State the ILATE rule and write the Integration by Parts formula.

   Priority order for choosing u: $\text{Inverse trig} > \text{Logarithmic} > \text{Algebraic} > \text{Trigonometric} > \text{Exponential}$; Integration by Parts: $\int u\,v\,dx = u\int v\,dx - \int\left(\dfrac{du}{dx}\int v\,dx\right)dx$

   Hint: Choose $u$ as the function higher in this order (ILATE).

3. 3.Evaluate $\int \frac{dx}{\sqrt{a^2-x^2}}$ and $\int \frac{dx}{\sqrt{x^2-a^2}}$.

   $\sin^{-1}\left(\dfrac{x}{a}\right)+C$ and $\ln\left|x+\sqrt{x^2-a^2}\right|+C$

   Hint: Standard inverse trig / log forms.

4. 4.Write the reduction technique used to evaluate $\int \frac{dx}{a\sin x + b\cos x + c}$ (type of substitution).

   Use the Weierstrass (half-angle) substitution $t=\tan\left(\dfrac{x}{2}\right)$, giving $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$, converting the integral into a rational function of $t$ integrable by standard methods.

   Hint: Half-angle tangent substitution.

5. 5.For evaluating $\int \frac{px+q}{ax^2+bx+c}dx$, how is the numerator split?

   Write $px+q = \lambda\left(2ax+b\right)+\mu$, where $2ax+b$ is the derivative of the denominator; comparing coefficients gives $\lambda,\mu$, splitting the integral into a logarithmic term (from $\lambda$) and a standard integral term (from $\mu$).

   Hint: Express numerator as derivative of denominator plus a constant.

6. 6.Evaluate $\int e^x\sec x(1+\tan x)\,dx$.

   Write $\sec x(1+\tan x)=\sec x+\sec x\tan x=f(x)+f'(x)$ with $f(x)=\sec x$. Since $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+C$, the integral equals $e^x\sec x+C$.

   Hint: Spot the $e^x[f(x)+f'(x)]$ pattern with $f=\sec x$.

7. 7.Evaluate $\int x\sec^2x\,dx$.

   Integrate by parts with $u=x,\,dv=\sec^2x\,dx$, so $du=dx,\,v=\tan x$: $\int x\sec^2x\,dx=x\tan x-\int\tan x\,dx=x\tan x+\ln|\cos x|+C$.

   Hint: Take $u=x$ and integrate $\sec^2x$ first.

8. 8.Evaluate $\int e^{ax}\cos(bx)\,dx$ by repeated integration by parts.

   Applying integration by parts twice returns the original integral, giving $\int e^{ax}\cos(bx)\,dx\big(a^2+b^2\big)=e^{ax}(a\cos bx+b\sin bx)$. Hence $\int e^{ax}\cos(bx)\,dx=\dfrac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C$.

   Hint: Do IBP twice; the original integral reappears — solve algebraically.

9. 9.Evaluate $\int \ln\!\left(x+\sqrt{1+x^2}\right)dx$.

   With $u=\ln(x+\sqrt{1+x^2})$, $dv=dx$: $du=\dfrac{dx}{\sqrt{1+x^2}}$, $v=x$. So the integral is $x\ln(x+\sqrt{1+x^2})-\int\dfrac{x}{\sqrt{1+x^2}}\,dx=x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C$.

   Hint: IBP with $u=\sinh^{-1}x$, $dv=dx$.

10. 10.Derive $\int \sec^3x\,dx$ using integration by parts on $\sec x\cdot\sec^2x$.

    With $u=\sec x,dv=\sec^2x\,dx$: $\int\sec^3x\,dx=\sec x\tan x-\int\sec x\tan^2x\,dx=\sec x\tan x-\int\sec^3x\,dx+\int\sec x\,dx$. Solving, $2\int\sec^3x\,dx=\sec x\tan x+\ln|\sec x+\tan x|$, so $\int\sec^3x\,dx=\dfrac12\big[\sec x\tan x+\ln|\sec x+\tan x|\big]+C$.

    Hint: IBP gives back $\int\sec^3x\,dx$ on the right — solve for it.

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