MATH 1172 Midterm: MATH 1172 Ohio State University Math 1172 7.4 Solutions Fa 2014

• School: Ohio State University
• Department: Mathematics
• Course: MATH 1172
• Professor: All

Document Summary

After the above simpli cation we observe that the term under the radical is of the form a2 + x2, with a=2. Thus the substi- tution x = 2 tan , with dx = 2 sec2 d , will simplify the integral. 2 ln|sec + tan | + c. Now we just need to nd sec and tan in terms of x. We use the pythagorean theorem to show that the hypotenuse is 4 + x2. With this in mind sec = 4 + x2/2 and tan = x/2. Solution: the term under the radical is of the form variable squared less constant squared. 1 = tan2 , we should make the substitution x = 10 sec , with dx = 10 sec tan d . a dx dx x3 x2 100. 10 sec tan d tan d . When [0, /2] or x [1, ), we have that.