Evaluate the integral in terms of natural logarithms.
-$$\int _ { 0 } ^ { 11 \pi } \frac { - \sin x d x } { \sqrt { 1 + \cos ^ { 2 } x } }$$
A) $\ln 2$
B) $\ln \left( \frac { - 1 + \sqrt { 2 } } { 1 + \sqrt { 2 } } \right)$
C) $- 2$
D) 0

Question

Quizplus · Accepted Answer

We begin with the substitution $u = \cos x$ to transform the integral into
$$\int_{-1}^1 \frac{du}{\sqrt{2-u^2}} = \sin^{-1}\left(\frac{u}{\sqrt{2}}\right)\biggr\vert_{-1}^1 = \frac{\pi}{2}.$$
Therefore, the answer is $\ln\left(\frac{-1+\sqrt{2}}{1+\sqrt{2}}\right)$ by the formula $\int_a^b f(x) dx = F(b) - F(a),$ where $F(x)$ is the antiderivative of $f(x).$

Evaluate the Integral in Terms of Natural Logarithms ∫011π−sin⁡xdx1+cos⁡2x\int _ { 0 } ^ { 11 \pi } \frac { - \sin x d x } { \sqrt { 1 + \cos ^ { 2 } x } }∫011π​1+cos2x​−sinxdx​

Evaluate the Integral in Terms of Natural Logarithms $\int _ { 0 } ^ { 11 \pi } \frac { - \sin x d x } { \sqrt { 1 + \cos ^ { 2 } x } }$