Question Number 59563 by aliesam last updated on 11/May/19
Commented by aliesam last updated on 11/May/19
$${thank}\:{you}\:{sir} \\ $$
Answered by MJS last updated on 12/May/19
![∫((sin^(−7) x)/(cos^3 x))dx=∫(dx/(sin^7 x cos^3 x))= [t=sin x → dx=(dt/(cos x))] =∫(dt/(t^7 (1−t^2 )^2 ))=∫((1/(4(1−t)^2 ))−(1/(4(1+t)^2 ))+(2/(1−t))−(2/(1+t))+(4/t)+(3/t^3 )+(2/t^5 )+(1/t^7 ))dt and now it should be easy](https://www.tinkutara.com/question/Q59570.png)
$$\int\frac{\mathrm{sin}^{−\mathrm{7}} \:{x}}{\mathrm{cos}^{\mathrm{3}} \:{x}}{dx}=\int\frac{{dx}}{\mathrm{sin}^{\mathrm{7}} \:{x}\:\mathrm{cos}^{\mathrm{3}} \:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cos}\:{x}}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{7}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }=\int\left(\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{1}−{t}}−\frac{\mathrm{2}}{\mathrm{1}+{t}}+\frac{\mathrm{4}}{{t}}+\frac{\mathrm{3}}{{t}^{\mathrm{3}} }+\frac{\mathrm{2}}{{t}^{\mathrm{5}} }+\frac{\mathrm{1}}{{t}^{\mathrm{7}} }\right){dt} \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\ $$
Commented by malwaan last updated on 11/May/19
$${but}\:{your}\:{solution}\:{is}\:{very}\:{short} \\ $$$${Can}\:{you}\:{make}\:{steps}\:{please}? \\ $$
Answered by tanmay last updated on 12/May/19
![∫(dx/(sin^7 xcos^3 x)) ∫(dx/(tan^7 x×cos^(10) x)) ∫((sec^2 x×sec^8 x)/(tan^7 x))dx ∫(((1+t^2 )^4 dt)/t^7 ) [t=tanx] ∫((1+4c_1 t^2 +4c_2 t^4 +4c_3 t^6 +t^8 )/t^7 )dt ∫t^(−7) dt+4∫t^(−5) dt+6∫t^(−3) +4∫(dt/t)+∫tdt (t^(−6) /(−6))+4×(t^(−4) /(−4))+6×(t^(−2) /(−2))+4lnt+(t^2 /2)+c ((−1)/(6(tanx)^6 ))−(1/((tanx)^4 ))−(3/((tanx)^2 ))+4ln(tanx)+(((tanx)^2 )/2)+c](https://www.tinkutara.com/question/Q59584.png)
$$\int\frac{{dx}}{{sin}^{\mathrm{7}} {xcos}^{\mathrm{3}} {x}} \\ $$$$\int\frac{{dx}}{{tan}^{\mathrm{7}} {x}×{cos}^{\mathrm{10}} {x}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} {x}×{sec}^{\mathrm{8}} {x}}{{tan}^{\mathrm{7}} {x}}{dx} \\ $$$$\int\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} {dt}}{{t}^{\mathrm{7}} }\:\:\left[{t}={tanx}\right] \\ $$$$\int\frac{\mathrm{1}+\mathrm{4}{c}_{\mathrm{1}} {t}^{\mathrm{2}} +\mathrm{4}{c}_{\mathrm{2}} {t}^{\mathrm{4}} +\mathrm{4}{c}_{\mathrm{3}} {t}^{\mathrm{6}} +{t}^{\mathrm{8}} }{{t}^{\mathrm{7}} }{dt} \\ $$$$\int{t}^{−\mathrm{7}} {dt}+\mathrm{4}\int{t}^{−\mathrm{5}} {dt}+\mathrm{6}\int{t}^{−\mathrm{3}} +\mathrm{4}\int\frac{{dt}}{{t}}+\int{tdt} \\ $$$$\frac{{t}^{−\mathrm{6}} }{−\mathrm{6}}+\mathrm{4}×\frac{{t}^{−\mathrm{4}} }{−\mathrm{4}}+\mathrm{6}×\frac{{t}^{−\mathrm{2}} }{−\mathrm{2}}+\mathrm{4}{lnt}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$\frac{−\mathrm{1}}{\mathrm{6}\left({tanx}\right)^{\mathrm{6}} }−\frac{\mathrm{1}}{\left({tanx}\right)^{\mathrm{4}} }−\frac{\mathrm{3}}{\left({tanx}\right)^{\mathrm{2}} }+\mathrm{4}{ln}\left({tanx}\right)+\frac{\left({tanx}\right)^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$
Commented by malwaan last updated on 12/May/19
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}} \\ $$