Trigonometric Substitution Examples With Solutions Pdf

X45tanθ And the derivative becomes. Then and Note that because Thus the Inverse Substitution Rule gives cot C y csc2 1 d y cos2 sin2 d y cot2 d y s9 x2 x2 dx y 3 cos 9 sin2 3 cos d cos 0 2 2 s9 x2 s9 9 sin2 s 9 cos2 3 cos 3 cos x 3 sin 2 2 dx 3 cos d y s9 x2 x2 dx 2 2 x a sin y f x dx y f t t t t dt.

Gebhard Curt Trig Notes

932 sqrt93 33 27 This is a well-known trigonometric identity.

Trigonometric substitution examples with solutions pdf. D x 4 sec 2 u. In the traditional substitution we dene the new variable in terms of the old. Substitute into the original problem replacing all forms of getting Use antiderivative rule 2 from the beginning of this section.

Z 1 tan2 x sec2 x dx Z cos2 x sin2 xdx Z cos2xdx 1 2 sin2x C 18. Find displaystyle int frac1sqrt4x2 dx. 81 Substitution 167 then the integral becomes Z 2xcosx2dx Z 2xcosu du 2x Z cosudu.

For example u 1 x2. Then Z axbndx Z 1 a undu 1. The important thing to remember is that you must eliminate all instances of the original variable x.

Example Z x3 p 4 x2 dx I Let x 2sin dx 2cos d p 4x2 p 4sin2 2cos. Annette Pilkington Trigonometric Substitution. Well need to use the following.

We have Z sin5 xdx Z sin4 xsinxdx Z sin 2x2 sinxdx Z 1cos2 x sinxdx cosx u dcosx du sinxdx du sinxdx du Z 1u 2 du Z 12u2 u4du u 2 3 u3 1 5 u5 C cosx 2 3 cos3 x 1 5. Evaluate Z x2 4 2x32 dx. Multiply by the conjugate.

Opposite sin hypotenuse q hypotenuse csc opposite q adjacent cos hypotenuse q hypotenuse sec adjacent q opposite tan adjacent q adjacent cot opposite q Unit circle definition For this definition q is any. àFirst complete the square T v FT 6 v Note. Z dx cosx 1 Z 1 cosx 1 cosx 1 cosx 1 dx Z cosx 1 cos2 x 1 dx Z cosx 1 sin2 x dx Z.

Table of Trigonometric Substitutions EXAMPLE 1 Evaluate. We find Z sinx1cos2 xcos2 xdx Z 1 u2u2 du Z u4 u2du u5 5 u3 3 c 1 5 cos5 x 1 3 cos3 x c In the case when m is even and n is odd we can proceed in a similar fashion use the identity cos2 A 1 sin2 A and the substitution u sinx. Then du 3 2 dx and the integral becomes 1 4 Z 1 1 9 4 x2 dx 1 4 Z 1 1u2 2 3 du 1 6 Z 1 1u2 du This can be finished off using the standard result to give 1 6 tan1 uc 1 6 tan1 3 2 x c.

Drawing a reference triangle we nd that Z x2 4 3x2 2 dx tan C x p 4 x2 sin 1 x 2 C. Section 1-3. For problems 1 8 use a trig substitution to eliminate the root.

Then Z x 2 4 3x2 2 dx Z 4sin 2cos 8cos3 d Z sin2 cos2 d Z tan2 d Z sec2 1d tan C. SOLUTION Let where. Show that the translation is.

Click HERE to return to the list of problems. Int frac 1 16 4tan u 2times 4sec 2u du 164tanu21. Let x 2sin dx 2cos d 4 x 23 2cos 3 8cos.

Let so that or. Let us observe the effect of making the substitution u 3 2 x so that u2 9 4 x2. Let x2 tan theta so that dx 2sec2 theta dtheta and beginalign int frac1sqrt4x2 dx int frac2sec2 theta2sectheta dtheta int sectheta dtheta ln sec theta tantheta C ln leftfracxsqrtax22right C endalign where C is an arbitrary constant.

Dierence between the substitution technique learned before and the one we are about to explain. This is using the half identity backwards. Let x tan then dx sec2 d.

R 1 tan2 x sec2 x dx Solution. 2 dx45sec θdθ And the substitution that involves the integrand becomes. I Let w cos dw sin d 8 Z 1 cos2 sin d 8 Z 1 w2 dw 8 Z w2 1 dw 8w3 3 8wC 8cos 3 3 18cos C 8cossin 1 x 23 3 8cossin x 2 C.

Completing the Square 9. 49z2 4 9 z 2 Solution. Xatan u x atan u.

7t2 35 2 7 t 2 3 5 2 Solution. I R px 3dx 4 2x R 8sin 2cos d 2cos R 8sin3 d R 8sin2 sin d 8 R 1 cos2 sin d. Find Z sin5 xdx.

We now consider a similar example for which a sine substitution is appropriate. W32 100 w 3 2 100 Solution. MATH 142 - Trigonometric Substitution Joe Foster Example 2 Evaluate ˆ 1 x2 x2 4 dx.

EXAMPLE811 Evaluate Z axbndx assuming that a and b are constants a 6 0 and n is a positive integer. R dx cosx 1 Solution. 1 1 6 4 tan u 2 4 sec 2 u d u.

Let so that or. In trigonometric substitution we redene the given variable. A2 x2 1tan2 u sec2 u x atanu dx asec2 udu a2 x2 a2 sec2 u Example 9.

X 2tanθ θ π 2 π 2 dx 2sec2 θ dθ u sinθ du cosθ ˆ 1 x2 x2 4 dx ˆ 2sec2 θ 22 tan 2θ p 2 tan2 2 dθ ˆ 2sec 2 θ 22 tan2 θ 2 p tan2 θ 1 dθ ˆ sec2 θ 22 tan2 θ p sec2 θ dθ ˆ secθ 22 tan2 θ dθ 1 4 ˆ cosθ sin2 θ dθ 1 4 ˆ 1 u2 du 1 4 1 u C 1. Dx4sec 2u dx 4sec2 u. We make the first substitution and simplify the denominator of the question before proceeding to integrate.

Substituting the variables from the above we have. Example 1 Now we choose a proper translation from x to θ. Z p 1x2 x dx Z p 1tan2 tan sec2 d Z sec sec2 tan d Z sec 1tan2 tan d Z sec tan sec tan2 tan d Z csc sec tan d lnjcsc cot jsec C Our substitution was x tan so our triangle has opposite side x adjacent side 1 and hypotenuse p x2 1.

Heres a number example demonstrating this expression. If the integrand involves p. Then Z p 1x2 x dx lnjcsc cot jsec C ln p x2 1 x 1 x p x2 1C 5.

We let u ax b so du adx or dx dua. At this stage the substitution u cosx du sinxdx enables us to rapidly complete the solution. Section 62 Trigonometric Integrals and Substitutions 2010 Kiryl Tsishchanka EXAMPLE 2.

Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p. SOLUTIONS TO TRIGONOMETRIC INTEGRALS SOLUTION 1. 1325x2 13 25 x 2 Solution.

Z ˇ4 0 p 1 cos4 d Z ˇ4 0 r 2 1 2 1 cos4 d Z ˇ4 0 p 2sin2 2 d p 2 Z ˇ4 0 sin2 d p 2 1 2 cos2 ˇ4 0 p 2 2 17. Tan2 θ 1 sec2 θ So we have. Z x2 4x 2 dx Z 4tan 2u 44tan u 2sec2 udu Z 8tan usec2 u 4sec u du Z 8tan2 usec2 u 2secu du 4 Z tan2 usecudu finish as in Example 12 Section 83 substitute in terms of x Example.

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