Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 4681 by thachan last updated on 21/Feb/16

∫_4 ^(cos Π_4 ) (x^2 +3ax−5tan (x))dx

$$\int_{\mathrm{4}} ^{\mathrm{cos}\:\underset{\mathrm{4}} {\prod}} \left({x}^{\mathrm{2}} +\mathrm{3}{ax}−\mathrm{5tan}\:\left({x}\right)\right){dx} \\ $$

Commented by prakash jain last updated on 21/Feb/16

is the upper limit cos (π/4)?

$${is}\:{the}\:{upper}\:{limit}\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}? \\ $$

Commented by thachan last updated on 21/Feb/16

yes.it is cos

$${yes}.{it}\:{is}\:{cos} \\ $$

Answered by FilupSmith last updated on 22/Feb/16

=∫_4 ^( cos(π/4)) x^2 dx+3a∫_4 ^( cos(π/4)) xdx+5∫_4 ^( cos(π/4)) tan(x)dx  =(1/3)[(1/(2(√2)))−4^3 ]+(3/2)a[(1/2)−16]+5[ln(cos(4))−ln(cos((1/(√2))))]  =(((3−96)a)/4)−((63)/(6(√2)))+5ln(((cos 4)/(cos 2^(−0.5) )))  =((252−558a(√2))/(24(√2)))+5ln(((cos 4)/(cos 2^(−0.5) )))  continue from here

$$=\int_{\mathrm{4}} ^{\:\mathrm{cos}\left(\pi/\mathrm{4}\right)} {x}^{\mathrm{2}} {dx}+\mathrm{3}{a}\int_{\mathrm{4}} ^{\:\mathrm{cos}\left(\pi/\mathrm{4}\right)} {xdx}+\mathrm{5}\int_{\mathrm{4}} ^{\:\mathrm{cos}\left(\pi/\mathrm{4}\right)} \mathrm{tan}\left({x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}−\mathrm{4}^{\mathrm{3}} \right]+\frac{\mathrm{3}}{\mathrm{2}}{a}\left[\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{16}\right]+\mathrm{5}\left[\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{4}\right)\right)−\mathrm{ln}\left(\mathrm{cos}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right)\right] \\ $$$$=\frac{\left(\mathrm{3}−\mathrm{96}\right){a}}{\mathrm{4}}−\frac{\mathrm{63}}{\mathrm{6}\sqrt{\mathrm{2}}}+\mathrm{5ln}\left(\frac{\mathrm{cos}\:\mathrm{4}}{\mathrm{cos}\:\mathrm{2}^{−\mathrm{0}.\mathrm{5}} }\right) \\ $$$$=\frac{\mathrm{252}−\mathrm{558}{a}\sqrt{\mathrm{2}}}{\mathrm{24}\sqrt{\mathrm{2}}}+\mathrm{5ln}\left(\frac{\mathrm{cos}\:\mathrm{4}}{\mathrm{cos}\:\mathrm{2}^{−\mathrm{0}.\mathrm{5}} }\right) \\ $$$$\mathrm{continue}\:\mathrm{from}\:\mathrm{here} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com