# JEE Main & Advanced Mathematics Definite Integrals Some Important Results of Definite Integral

Some Important Results of Definite Integral

Category : JEE Main & Advanced

(1) If ${{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}xdx}$ then ${{I}_{n}}+{{I}_{n-2}}=\frac{1}{n-1}$

(2) If ${{I}_{n}}=\int_{0}^{\pi /4}{{{\cot }^{n}}xdx}$ then ${{I}_{n}}+{{I}_{n-2}}=\frac{1}{1-n}$

(3) If ${{I}_{n}}=\int_{0}^{\pi /4}{{{\sec }^{n}}x\,dx}$ then ${{I}_{n}}=\frac{{{(\sqrt{2})}^{n-2}}}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}$

(4) If ${{I}_{n}}=\int_{0}^{\pi /4}{\text{cose}{{\text{c}}^{\text{n}}}x\,dx}$ then ${{I}_{n}}=\frac{{{(\sqrt{2})}^{n-2}}}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}$

(5) If ${{I}_{n}}=\int_{0}^{\pi /2}{{{\sin }^{n}}x\,}dx,\,$ then ${{I}_{n}}=\frac{n-1}{n}{{I}_{n-2}}$

(6) If ${{I}_{n}}=\int_{0}^{\pi /2}{{{\cos }^{n}}x\,dx,}$ then ${{I}_{n}}=\frac{n-1}{n}{{I}_{n-2}}$.

(7) If ${{I}_{n}}=\int_{0}^{\pi /2}{{{x}^{n}}\sin x\,dx}$ then ${{I}_{n}}+n(n-1){{I}_{n-2}}=n{{(\pi /2)}^{n-1}}$

(8) If ${{I}_{n}}=\int_{0}^{\pi /2}{{{x}^{n}}\cos x\,dx}$ then ${{I}_{n}}+n(n-1){{I}_{n-2}}=\,{{(\pi /2)}^{n}}$

(9)  If $a>b>0,$ then $\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x}}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}{{\tan }^{-1}}\sqrt{\frac{a+b}{a-b}}$

(10)If $n\,\in \,I$ then $\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x}}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\log \left| \frac{\sqrt{b+a}-\sqrt{b-a}}{\sqrt{b+a}+\sqrt{b-a}} \right|$

(11) If $a>b>0$ then $\int_{0}^{\pi /2}{\frac{dx}{a+b\sin x}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}{{\tan }^{-1}}\sqrt{\frac{a-b}{a+b}}}$

(12) If $0<a<b$, then

$\int_{0}^{\pi /2}{\frac{dx}{a+b\sin x}}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\log \left| \frac{\sqrt{b+a}+\sqrt{b-a}}{\sqrt{b+a}-\sqrt{b-a}} \right|$

(13) If $a>b,\,{{a}^{2}}>{{b}^{2}}+{{c}^{2}},$ then $\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}$$\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}}}\,{{\tan }^{-1}}\frac{a-b+c}{\sqrt{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}}}$

(14) If $a>b,\,{{a}^{2}}<{{b}^{2}}+{{c}^{2}},$ then $\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}$ $=\frac{1}{\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}$ $\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}=\frac{1}{\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}}\log \left| \frac{a-b+c-\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}{a-b+c+\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}} \right|$

(15) If $a<b,$ ${{a}^{2}}<{{b}^{2}}+{{c}^{2}}$ then $\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}$$=\frac{-1}{\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}\log \left| \frac{b-a-c-\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}{b-a-c+\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}} \right|$.

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