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Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

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Sec. 2] Integration by Substitution<br />

Solution. Put xlant. Therefore, dx= r-r ,<br />

cos 2 /<br />

f y~x*+l _ f V~ian 2 / + 1 dt f sec t cos' i dt 1?<br />

*<br />

J tan<br />

_<br />

2 / cos ! < J s<strong>in</strong> 2 .-cos /<br />

2 /<br />

dt<br />

f<br />

f si " 2 * 4- cos2 * M- ^ ^/ "~<br />

J<br />

,<br />

1<br />

f<br />

s<strong>in</strong> 2 ~~<br />

/ cos / J s<strong>in</strong> 2 ~<br />

/-cos/ J cos / J s<strong>in</strong> 2 /<br />

= In tan /<br />

| + sec /<br />

1<br />

-J + C = 2<br />

In / -{- V\ -j-tan /<br />

1<br />

|<br />

J<br />

tan<<br />

tan<br />

cos/<br />

~~ _<br />

1191. Apply<strong>in</strong>g the <strong>in</strong>dicated substitutions, f<strong>in</strong>d the follow<strong>in</strong>g<br />

<strong>in</strong>tegrals:<br />

c)<br />

f<br />

e) \<br />

J<br />

COS A' d*<br />

Apply<strong>in</strong>g suitable substitutions, f<strong>in</strong>d the follow<strong>in</strong>g <strong>in</strong>tegrals:<br />

1192. S*(2x+5)"djc.<br />

1193. ('<br />

1 + *..d*. 1198.<br />

J l+^A-<br />

1194. f- J!<br />

JxK2t+l ,, 99<<br />

**<br />

1195. r<br />

.<br />

>97. n- cs<strong>in</strong>A )"<br />

Apply<strong>in</strong>g trigonometric substitutions, f<strong>in</strong>d the follow<strong>in</strong>g <strong>in</strong>tegrals:<br />

,201. ('-=*. 1203.<br />

J K l *' J ^<br />

,202. -=. 1204*.<br />

f iZEl'dx.<br />

1J5

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