Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 44575 by Raj Singh last updated on 01/Oct/18

Commented by $@ty@m last updated on 01/Oct/18

similar to Q. No. 41703

$${similar}\:{to}\:{Q}.\:{No}.\:\mathrm{41703} \\ $$

Commented by Raj Singh last updated on 02/Oct/18

this is sin2Θ

$${this}\:{is}\:{sin}\mathrm{2}\Theta \\ $$

Commented by $@ty@m last updated on 02/Oct/18

it doesn′t matter. sin 2θ=2sin θcos θ when divided by cos^4 θ, N^r =2tan θ.sec^2 θ .... and then the procedure will be similar. .... our ultimate aim is to get tan θ .

$${it}\:{doesn}'{t}\:{matter}. \\ $$$$\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2sin}\:\theta\mathrm{cos}\:\theta \\ $$$${when}\:{divided}\:{by}\:\mathrm{cos}\:^{\mathrm{4}} \theta, \\ $$$${N}^{{r}} =\mathrm{2tan}\:\theta.\mathrm{sec}\:^{\mathrm{2}} \theta \\ $$$$....\:{and}\:{then}\:{the}\:{procedure}\:{will}\: \\ $$$${be}\:{similar}. \\ $$$$....\:{our}\:{ultimate}\:{aim}\:{is}\:{to}\:{get}\:\mathrm{tan}\:\theta\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Oct/18

∫_0 ^(π/4) (((sin^2 θ)/(cos^4 θ))/(1+tan^4 θ))dθ ∫_0 ^(π/4) ((tan^2 θsec^2 θ)/(1+tan^4 θ))dθ t=tanθ dt=sec^2 θ dθ ∫_0 ^1 ((t^2 ×dt)/(1+t^4 )) ∫_0 ^1 (dt/(t^2 +(1/t^2 ))) (1/2)∫_0 ^1 ((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt (1/2)∫_0 ^1 ((d(t+(1/t)))/((t+(1/t))^2 −2))+(1/2)∫_0 ^1 ((d(t−(1/t)))/((t−(1/t))^2 +2)) (1/2){(1/(2(√2) ))∣ln(((t+(1/t)−(√2))/(t+(1/t)+(√2))))∣_0 ^1 +(1/(√2))∣tan^(−1) (((t−(1/t))/(√2)))∣_0 ^1 } (1/2){(1/(2(√2)))∣ln(((t^2 +1−(√2))/(t^2 +1+(√(2 )))))∣_0 ^1 +(1/(√2))∣tan^(−1) (((t^2 −1)/(√2)))∣_0 ^1 } (1/2){(1/(2(√2)))∣ln(((2−(√2))/(2+(√2))))−ln(((1−(√2))/(1+(√2))))∣+(1/(√2))tan^(−1) (0)−tan^(−1) (((−1)/(√2)))} (1/2){(1/(2(√2)))∣ln((((√2) −1)/((√2) +1)))−ln(((1−(√2))/(1+(√2))))+(1/(√2))(tan^(−1) ((1/(√2)))}

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{4}} \theta}}{\mathrm{1}+{tan}^{\mathrm{4}} \theta}{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{tan}^{\mathrm{2}} \theta{sec}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{4}} \theta}{d}\theta \\ $$$${t}={tan}\theta\:\:\:{dt}={sec}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} ×{dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}\mid{ln}\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mid{ln}\left(\frac{{t}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}\:}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mid{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)−{ln}\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}}\right)\mid+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)−{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mid{ln}\left(\frac{\sqrt{\mathrm{2}}\:−\mathrm{1}}{\sqrt{\mathrm{2}}\:+\mathrm{1}}\right)−{ln}\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right\}\right. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com