Chapter 1

\(\int 0 d x=\) constant

STANDARD INTEGRALS \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\) or \(-\cos ^{-1}\left(\frac{x}{a}\right)\)

$$ \int \frac{d t}{\sqrt{(t+\alpha)(t+\beta)}}=2 \log [\sqrt{t+\alpha}+\sqrt{t+\beta}] $$

Fvaluate \(\int \frac{e^{x}\left(x^{2}+1\right) d x}{(x+1)^{2}}\)

Fvaluate \(\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x\)

$$ \int \frac{d x}{\sqrt{a^{2}-b^{2} x^{2}}}=\frac{1}{b} \sin ^{-1} \frac{b x}{a} $$

$$ \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right| \text { or } \frac{1}{a} \tan h^{-1}\left(\frac{x}{a}\right) $$

Lvaluate \(\int \sqrt{\frac{a+x}{x}} d x\)

Integrals of the form: \(\int \cot ^{m} x \operatorname{cosec}^{n} x d x\)Rule (i) If \(m\) is even or odd integer and \(n\) is even positive integer then put \(\cot x=t\). (ii) If \(m\) is odd positive integer and \(n \notin\) even positive integer then put cosec \(x=t\).(iii) I \(m=0\) and \(n=2 r+1 \forall r \in N\), then \(\int \operatorname{cosec}^{2 r+1} x d x=\int \operatorname{cosec}^{2 r-1} x \operatorname{cosec}^{2} x d x\), then integrate by parts taking \(\operatorname{cosec}^{2} x\) as second function.

\(\int \frac{d x}{(a x+b)^{2}}=\frac{-1}{a(a x+b)}\)

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

Lvaluate \(\int e^{x \mathrm{v}} \cdot \sin 2 x d x\)

$$ \begin{gathered} \sec ^{-1} x+c \\ \int \frac{1}{x \sqrt{x^{2}-1}} d x= \\ -\operatorname{coscc}^{-1} x+c,|x|>1 \end{gathered} \text { or } $$

Fvaluate \(\int \frac{x}{\sqrt{8+x-x^{2}}} d x\)

\(\int \frac{d x}{\sqrt{a x+b}}=\frac{2 \sqrt{a x+b}}{a}\)

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

Fvaluate \(\int \frac{1}{(2 x+3) \sqrt{x+1}} d x\)

\(\int \frac{d x}{c^{2}+(a x+b)^{2}}=\frac{1}{a c} \tan ^{-1} \frac{a x+b}{c}\)

Lvaluate \(\int \frac{1}{(x+1) \sqrt{x^{3}-1}} d x\)

\(\int \frac{d x}{\left(x^{2}+k\right)^{n}}=\frac{x}{k(2 n-2)\left(x^{2}+k\right)^{n-1}}+\frac{2 n-3}{k(2 n-2)}\) \(\int \frac{d x}{\left(x^{2}+k\right)^{n-1}}\)

\(\int e^{x} d x=e^{x}+c\)

$$ \begin{aligned} &\int \frac{d x}{\cos (x-a) \cos (x-b)}\\\ &=\frac{1}{\sin (a-b)} \log \frac{\cos (x-a)}{\cos (x-b)}\\\ &\text { C. TM commit to memory. } \end{aligned} $$

Integrals of the Form (a) \(\int \frac{d x}{a x^{2}+b x+c}\) (b) \(\int \frac{d x}{\sqrt{a x^{2}+b x+c}}\) (c) \(\int \sqrt{a x^{2}+b x+c} d x\)Working Rule (i) Make the cocllicient of \(x^{2}\) unity by taking the coclficicnt of \(x^{2}\) outside the quadratic. (ii) Complete the square in the terms involving \(x\), i. \(.\), write \(\alpha x^{2}+b x+c\) in the form \(a[x \pm\) \(\left.\alpha)^{2} \pm \beta^{3}\right]\) (iii) The integrand is converted to one of the nine special integrals. (iv) Integrate the function.

\(\int \sin x d x=-\cos x+c\)

\(\int \frac{d x}{x \sqrt{x^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1}\left(\frac{x}{a}\right)\)

$$ \int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\log \left(x+\sqrt{x^{2}-a^{2}}\right] \text { or } \cos h^{-1}\left(\frac{x}{a}\right) $$

Integrals of the form (a) \(\int \frac{p x+q}{a x^{2}+b x+c} d x\)Working Rule $$ \text { Put } p x+q=\lambda(2 a x+b)+\mu \text { or } $$ \(p x+q=\lambda\) (derivative of quadratic) \(+\mu\). Comparing the coefficient of \(x\) and constant term on both sides, we get $$ \begin{aligned} &p=2 a \lambda \text { and } q=b \lambda+\mu \Rightarrow \lambda=\frac{p}{2 a} \\\ &\text { and } \mu=\left(q-\frac{b p}{2 a}\right) \end{aligned} $$ Then the integral becomes $$ =\frac{p}{2 a} \int \frac{2 a x+b}{a x^{2}+b x+c} d x+\left(q-\frac{b p}{2 a}\right) \int \frac{d x}{a x^{2}+b x+c} $$ $$ =\frac{p}{2 a} \log \left|a x^{2}+b x+c\right|+\left(q-\frac{b p}{2 a}\right) \int \frac{d x}{a x^{2}+b x+c} $$

\(\int \cos x d x=\sin x+c\)

Integral of the Form (a) \(\int \frac{d x}{a+b \cos x}\) (b) \(\int \frac{d x}{a+b \sin x}\) (c) \(\int \frac{d x}{a+b \cos x+c \sin x}\) Working Rule (i) \(\mathrm{P}\) ut \(\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\) and \(\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\) so that the given integrand becomes a function of \(\tan \frac{x}{2}\) (ii) Put \(\tan \frac{x}{2}=t=>\frac{1}{2} \operatorname{scc}^{2} \frac{x}{2} d x=d t\) (iii) Integrate the resulting rational algebraic lunction of \(t\) (iv) In the answer, put \(t=\tan \frac{x}{2}\).

\(\int \tan x d x=\log |\operatorname{scc} x|+c\) or \(-\log |\cos x|\)

\(\int \frac{a d-b c}{(c x+d)^{2}} d x=\frac{a x+b}{c x+d}\)

. Intcgrals of the Form \(\int \frac{a \cos x+b \sin x}{c \cos x+d \sin x} d x\) Working Rule (i) Put Numerator: \(\lambda\) (denominator) \(+\mu\) (derivative of denominator) \(a \cos x+b \sin x=\lambda(c \cos x+d \sin x)+\) \(\mu(-c \sin x+d \cos x)\) (ii) Fquate cocfficients of \(\sin x\) and \(\cos x\) on both sides and find the values of \(\lambda\) and \(\mu\). (iii) Split the given integral into two integrals and cvaluate cach integral separately, i.c. \(\int \frac{a \cos x+b \sin x}{c \cos x+d \sin x} d x=\lambda\) \(\int 1 d x+\mu \int \frac{-c \sin x+d \cos x}{c \cos x+d \sin x} d x\) \(=\lambda x+\mu \log |a \cos x+b \sin x| .\) (iv) Substitute the values of \(\lambda\) and \(\mu\) found in step 2 .

\(\int \cot x d x=\log |\sin x|+c\)

\(\int \operatorname{coscc} x d x=\log (\operatorname{cosec} x-\cot x)=\log \tan \frac{x}{2}\)

\(\int x^{x}\left(1+\log _{e} x\right) d x=x^{x}\)

\(\int \operatorname{scc}^{2} x d x=\tan x+c\)

\(\int \frac{d x}{a+b \tan x}=\int \frac{\cos x d x}{a \cos x+b \sin x}\)

\(\int \operatorname{cosec}^{2} x d x=-\cot x+c\)

\(\int \frac{d x}{a+b \cot x}=\int \frac{\sin x d x}{a \sin x+b \cos x}=\)

\(\int \operatorname{scc} x \tan x d x=\sec x+c\)

\(\int \sec x \operatorname{cosec} x d x=\log \tan x+c\)

sIn first integral on R.H.S. put \(\tan x=t\) and in sceond integral put \(\sqrt{2} \sin x=u\) 1.e., \(\sec ^{2} d x=d t\) and \(\cos x d x=\frac{d u}{\sqrt{2}}\) \(\therefore I=\int \frac{d t}{\sqrt{\left(t^{2}-1\right)}}-\frac{2}{\sqrt{2}} \int \frac{d u}{\sqrt{\left(u^{2}-1\right)}}\) \(=\ln \left|t+\sqrt{\left(t^{2}-1\right)}\right|-\sqrt{2} \ln \left|u+\sqrt{\left(u^{2}-1\right)}\right|+c\) \(=\ln \left|\tan x+\sqrt{\left(\tan ^{2} x-1\right)}\right|-\sqrt{2} \ln \mid \sqrt{2} \sin\) \(x+\sqrt{\left(2 \sin ^{2} x-1\right)} \mid+c\) (ii) Evaluate \(\int \sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)} d x\) Solution Lct \(I=\int \sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)} d x=\int \frac{\left(\operatorname{cosec}^{2} x \pm a\right)}{\sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)}} d x\) \(=\int \frac{\operatorname{cosec}^{2} x d x}{\sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)}} \pm a \int \frac{d x}{\sqrt{\left(\operatorname{cosec}^{3} x \pm a\right)}}\) \(=\int \frac{\operatorname{coscc}^{2} x d x}{\sqrt{(1 \pm a)}+\cot ^{2} x} \pm a \int \frac{\sin x d x}{\sqrt{1 \pm a\left(1-\cos ^{2} x\right)}}\) In first integral on R.H.S. put \(\cot x=t\) and in second integral put \(\cos x=u\). (iii) Evaluate \(\int \sqrt{\left(\operatorname{coscc}^{2} x+3\right)} d x\)omc Important Questions with Solution: (i) Evaluate \(\int \sqrt{\left(\operatorname{scc}^{2} x-2\right)} d x\) Solution $$ \begin{aligned} \text { Let } I &=\int \sqrt{\left(\sec ^{2} x-2\right)} d x=\int \frac{\left(\sec ^{2} x-2\right)}{\sqrt{\left(\sec ^{2} x-2\right)}} d x \\ &=\int \frac{\sec ^{2} x d x}{\sqrt{\left(\sec ^{2} x-2\right)}}-2 \int \frac{d x}{\sqrt{\left(\sec ^{2} x-2\right)}} \\ &=\int \frac{\sec ^{2} x d x}{\sqrt{\left(\tan ^{2} x-1\right)}}-2 \int \frac{\cos x d x}{\sqrt{1-2\left(1-\sin ^{2} x\right)}} \\ &=\int \frac{\sec ^{2} x d x}{\sqrt{\left(\tan ^{2} x-1\right)}}-2 \int \frac{\cos x d x}{\sqrt{(\sqrt{2} \sin x)^{2}-1}} \end{aligned} $$

\(\int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left[x+\sqrt{x^{2}+a^{2}}\right]\) or \(\sin h^{-1}\left(\frac{x}{a}\right)\)