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

This preview shows page 71 - 74 out of 210 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Applied Calculus
The document you are viewing contains questions related to this textbook.
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!
The document you are viewing contains questions related to this textbook.
Applied Calculus
The document you are viewing contains questions related to this textbook.
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.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture