Integrals of Some Particular Functions

Integrals of the form:

$\int \frac{px+q}{\left(a{x}^2+bx+c\right)}dx,\int \frac{px+q}{\sqrt{a{x}^2+bx+c}}dx,\int \left(px+q\right)\sqrt{a{x}^2+bx+c}dx$

In these type of integrals we write px + q = $\ell$(diff coefficient of ax² + bx + c) + m.

Find$\ell$ and m by comparing the coefficient of x and constant term on both sides of the identity. In this way the question will reduce to the sum of two integrals which can be integrated easily.

Integral of the type $\frac{a{x}^2+bx+c}{p{x}^2+qx+r}or\frac{a{x}^2+bx+c}{\sqrt{p{x}^2+qx+r}}$

In this case substitute ax² + bx + c = M (px² + qx + r) + N (2px + q) + R

Find M, N, & R. The integration reduces to integration of three independent functions.

Illustration -1: Evaluate $\int \frac{x+1}{\sqrt{2x^2+x-3}}dx$.

Solution: Let x + 1 = (differential Coefficient of 2x² + x – 3) + B

x +1 = A(4x + 1) + B = 4ax + A + B

equating the coefficients, A =$\frac{1}{4}$, B = $\frac{3}{4}$

Now I = $\int \frac{x+1}{\sqrt{2x^2+x-3}}dx$

= $\frac{1}{4}\int \frac{4x+1}{\sqrt{2x^2+x-3}}dx+\frac{3}{4}\int \frac{dx}{\sqrt{2x^2+x-3}}$

Let I₁ = $\frac{1}{4}\int \frac{4x+1}{\sqrt{2x^2+x-3}}$ and I₂ = $\frac{3}{4}\int \frac{dx}{\sqrt{2x^2+x-3}}$

Put 2x² + x-3 =z $\Rightarrow$ (4x +1) dx =dz

I₁ = $\frac{1}{4}\int \frac{dz}{\sqrt{z}}$ = $\frac{1}{4}2\sqrt{z}$ = $\frac{1}{2}\sqrt{2x^2+x-3}$

I₂ = $\frac{3}{4}\int \frac{dx}{\sqrt{2}\sqrt{x^2+\frac{1}{2}x-\frac{3}{2}}}$ = $\frac{3}{4\sqrt{2}}\frac{dx}{\sqrt{{\left(x+\frac{1}{4}\right)}^2-\frac{25}{16}}}$

= $\frac{3}{4\sqrt{2}}\log \left[x+\frac{1}{4}+\sqrt{{\left(x+\frac{1}{4}\right)}^2-\frac{25}{16}}\right]+c$

Hence I = $\frac{1}{2}\sqrt{2x^2+x-3}+\frac{3}{4\sqrt{2}}\log \left(x+\frac{1}{4}+\sqrt{x+\frac{x}{2}-\frac{3}{2}}\right)+c$ .

Integration of Irrational Algebraic Fractions:

1. Irrational functions of (ax+b)^1/n and x can be easily evaluated by the substitution

tⁿ = ax+ b. Thus $\int f(x,{(ax+b)}^{1/n})dx=\int f\left(\frac{t^n-b}{a},t\right)\frac{n{t}^{n-1}}{a}dt$.

2. ${\int }_{}^{}\frac{dx}{{\left(x-k\right)}^r\sqrt{a{x}^2+bx+c}}$ . Here we substitute, x –k = 1/t.

This substitution will reduce the given integral to -${\int }_{}^{}\frac{t^{r-1}dt}{\sqrt{A{t}^2+Bt+c}}$.

3. $\text{To integrate }\frac{\text{1}}{\text{(A}{\text{x}}^{\text{2}}+B)\sqrt{C{x}^2+D}}\text{, we first put x = 1/t}$, so that

$\int \frac{\text{dx}}{\text{(A}{\text{x}}^{\text{2}}+B)\sqrt{C{x}^2+D}}=\int \frac{-1/t^{2}dt}{(A/t^2+B)\sqrt{C/t^2+D}}=-\int \frac{tdt}{(A+B{t}^2)\sqrt{C+D}{t}^2}$.

Now the substitution C + Dt² = u² reduces it to the form $\int \frac{\text{du}}{{\text{u}}^{\text{2}}\pm a^2}$.

4. $\int \frac{\left(a{x}^2+bx+c\right)dx}{\left(dx+e\right)\sqrt{f{x}^2+gx+h}}$

Here, we write, ax² +bx +c = A₁ (dx +e) ( 2fx +g) +B₁( dx +e) +C₁

where A₁, B₁ and C₁ are constants which can be obtained by comparing the coefficient of like terms on both sides. And given integral will reduce to the form

A₁$\int \frac{\left(2fx+g\right)}{\sqrt{f{x}^2+gx+}h}dx+B_1\int \frac{dx}{\sqrt{f{x}^2+gx+h}}+C_1\int \frac{dx}{\left(dx+e\right)\sqrt{f{x}^2+gx+h}}$

Illustration -2: Evaluate $\int \frac{\text{x}}{\text{(x-3)}\sqrt{\text{x+1}}}dx$.

Solution: Put x + 1 = t², we get $I=\int \frac{\text{(}{\text{t}}^{\text{2}}-1)2tdt}{(t^2-4)t}=2\int \frac{t^2-1}{t^2-4}dt$

$=2\int \left\{1+\frac{3}{t^2-4}\right\}dt=2t+\frac{3}{2}\ln \left|\frac{t-2}{t+2}\right|+c$ $=2\sqrt{x+1}+\frac{3}{2}\ln \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+c$

TRIGONOMETRIC INTEGRALS

Integrals of the form $\int$ R ( sinx, cosx) dx:

Here R is a rational function of sin x and cos x. This can be translated into integrals of a rational function by the substitution: tan(x/2) = t. This is the so called universal substitution. In this case

$\sin x\text{ = }\frac{\text{2t}}{\text{1+}{\text{t}}^{\text{2}}};\cos x=\frac{1-t^2}{1+t^2},x=\text{2 ta}{\text{n}}^{\text{-1}}t;dx=\frac{2dt}{1+t^2}$.

Sometimes, instead of the substitution tanx/2 = t, it is more advantageous to make the substitution cot x/2 = t

Universal substitution often leads to very cumbersome calculations. Indicated below are those cases where the aim can be achieved with the aid of simpler substitutions.

(a) If R(-sin x, cos x) = -R(sin x, cos x), substitute cos x = t

(b) If R(sin x, -cos x) = -R(sin x, cos x), substitute sin x = t

(c) If R(-sin x, - cos x) = R(sin x, cos x), substitute tan x = t

Integrals of the form:

$(i)\int \frac{p\text{ cosx + q sinx + r}}{\text{ a cosx+ b sinx + c}}dx(ii)\int \frac{p\text{ cosx + q sinx }}{\text{ a cosx+ b sinx }}dx$

Rule for (i) : In this integral express numerator as l (Denominator) + m(d.c. of denominator) + n. Find l, m, n by comparing the coefficients of sinx, cosx and constant term and split the integral into sum of three integrals.

$\ell \int dx\text{ + m }\int \frac{\text{d}\text{.c}\text{. of (Denominator)}}{\text{Denominator}}dx+n\int \frac{dx}{a\text{ cosx + b sinx + c}}$

Rule for (ii) : Express numerator as l (denominator) + m(d.c. of denominator) and find l and m as above

Illustration -3: Evaluate $\int \frac{dx}{\sin x\left(2{\cos }^{2}x-1\right)}$.

Solution: If in expression $\frac{1}{\sin x(2{\cos }^{2}x-1)}$ we substitute -sinx for sin x, then the integrand will change its sign. Hence, we take advantage of the substitution

t = cosx; dt = - sinx dx. This gives

\begin{align}  I=-\int{\dfrac{dt}{(1-{{t}^{2}})(2{{t}^{2}}-1)}},Since\,\dfrac{1}{(1-{{t}^{2}})(2{{t}^{2}}-1)}=\dfrac{2}{(1-2{{t}^{2}})}-\dfrac{1}{1-{{t}^{2}}} \\  \text{ I =}\dfrac{\text{-1}}{\sqrt{2}}\,\ln \,\left| \dfrac{1+\sqrt{2}\cos x}{1-\sqrt{2}\cos x} \right|+\dfrac{1}{2}\,\,\ln \,\left| \dfrac{1-\cos x}{1+\cos x} \right|+c \\ \end{align}

Integration of the Type: ò(sin^Mxcos^Nx)dx

M & N $\in$natural numbers.

If one of them is odd, then substitute for term of even power.

If both are odd, substitute either of the term.

If both are even, use trigonometric identities only.

Illustration -4: Evaluate $\int \sqrt{\sin x}\cos xdx$.

Solution: Put sinx = t

$\int t^{1/2}dt=\frac{2}{3}{t}^{3/2}+c$ = $\frac{2}{3}{\left(\sin x\right)}^{3/2}+c$.