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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