# Cosh 1 x d t d ln sinh t c cosh t d ln x c p x 2 1

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Chapter 6 / Exercise 72
Applied Calculus
Berresford/Rockett
Expert Verified
cosh 1 x D t D ln . sinh t C cosh t/ D ln .x C p x 2 1/: Because cosh t 1 for all t , this last expression is only valid for x D cosh t 1 . 32. Verify that tanh 1 x D 1 2 ln ˇ ˇ ˇ ˇ 1 C x 1 x ˇ ˇ ˇ ˇ for j x j < 1 . SOLUTION Let A D tanh 1 x . Then x D tanh A D sinh A cosh A D e A e A e A C e A :
##### We have textbook solutions for you!
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Chapter 6 / Exercise 72
Applied Calculus
Berresford/Rockett
Expert Verified
852 C H A P T E R 7 TECHNIQUES OF INTEGRATION Solving for A yields A D 1 2 ln x C 1 1 x I hence, tanh 1 x D 1 2 ln x C 1 1 x D 1 2 ln ˇ ˇ ˇ ˇ 1 C x 1 x ˇ ˇ ˇ ˇ : for j x j < 1 (so that both 1 C x and 1 x are positive). 33. Evaluate Z p x 2 C 16 dx using trigonometric substitution. Then use Exercise 31 to verify that your answer agrees with the answer in Example 3. SOLUTION Let x D 4 tan . Then dx D 4 sec 2 d and Z p x 2 C 16 dx D 16 Z sec 3 d D 8 tan sec C 8 Z sec d D 8 tan sec C 8 ln j sec C tan j C C D 8 x 4 p x 2 C 16 4 C 8 ln ˇ ˇ ˇ ˇ ˇ p x 2 C 16 4 C x 4 ˇ ˇ ˇ ˇ ˇ C C D 1 2 x p x 2 C 16 C 8 ln ˇ ˇ ˇ ˇ ˇ x 4 C r x 4 2 C 1 ˇ ˇ ˇ ˇ ˇ C C: Using Exercise 31, ln ˇ ˇ ˇ ˇ ˇ x 4 C r x 4 2 C 1 ˇ ˇ ˇ ˇ ˇ D sinh 1 x 4 ; so we can write the antiderivative as 1 2 x p x 2 C 16 C 8 sinh 1 x 4 C C; which agrees with the answer in Example 3. 34. Evaluate Z p x 2 9 dx in two ways: using trigonometric substitution and using hyperbolic substitution. Then use Exercise 31 to verify that the two answers agree. SOLUTION First, let x D 3 sec . Then dx D 3 sec tan d and Z p x 2 9 dx D 9 Z tan 2 sec d D 9 Z sec 3 d 9 Z sec d D 9 2 sec tan C 9 2 Z sec d 9 Z sec d D 9 2 sec tan 9 2 ln j sec C tan j C C D 9 2 x 3 p x 2 9 3 9 2 ln ˇ ˇ ˇ ˇ ˇ x 3 C p x 2 9 3 ˇ ˇ ˇ ˇ ˇ C C D 1 2 x p x 2 9 9 2 ln ˇ ˇ ˇ ˇ ˇ x 3 C r x 3 2 1 ˇ ˇ ˇ ˇ ˇ C C: Alternately, let x D 3 cosh t . Then dx D 3 sinh t dt and Z p x 2 9 dx D 9 Z sinh 2 t dt D 9 2 Z . cosh 2t 1/ dt D 9 2 sinh t cosh t 9 2 t C C D 1 2 x p x 2 9 9 2 cosh 1 x 3 C C: Using Exercise 31, cosh 1 x 3 D ln ˇ ˇ ˇ ˇ ˇ x 3 C r x 3 2 1 ˇ ˇ ˇ ˇ ˇ ; so our two answers agree.
S E C T I O N 7.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions 853 35. Prove the reduction formula for n 2 : Z cosh n x dx D 1 n cosh n 1 x sinh x C n 1 n Z cosh n 2 x dx 5 SOLUTION Using Integration by Parts with u D cosh n 1 x and v 0 D cosh x , we have Z cosh n x dx D cosh n 1 x sinh x .n 1/ Z cosh n 2 x sinh 2 x dx D cosh n 1 x sinh x .n 1/ Z cosh n x dx C .n 1/ Z cosh n 2 x dx: Adding .n 1/ R cosh n x dx to both sides then yields n Z cosh n x dx D cosh n 1 x sinh x C .n 1/ Z cosh n 2 x dx: Finally, Z cosh n x dx D 1 n cosh n 1 x sinh x C n 1 n Z cosh n 2 x dx: 36. Use Eq. (5) to evaluate Z cosh 4 x dx . SOLUTION Using Eq. (5) twice, Z cosh 4 x dx D 1 4 cosh 3 x sinh x C 3 4 Z cosh 2 x dx D 1 4 cosh 3 x sinh x C 3 4 1 2 cosh x sinh x C 1 2 Z dx D 1 4 cosh 3 x sinh x C 3 8 cosh x sinh x C 3 8 x C C: In Exercises 37–40, evaluate the integral.