Need Help?

  • Notes
  • Comments & Questions

Evaluate \(\int \frac{\tan ^{3}(x)}{\cos ^{3}(x)} d x\)

\(\int \frac{\tan ^{3}(x)}{\cos ^{3}(x)} d x=\int \tan ^{3}(x) \cdot \sec ^{3}(x) d x\)

\(\sec (x)=\frac{1}{\cos (x)}\)

\(\tan (x)=\frac{\cos (x)}{\cos (x)}\)

\(\int \frac{\tan ^{3}(x)}{\cos ^{3}(x)} d x=\int \frac{ \sin ^{3}(x)}{\cos ^{3}(x) \cdot \cos ^{3}(x)} d x=\int \frac{\sin ^{3} (x)}{\cos ^{6}(x)} d x\)

\(u=\cos x \rightarrow d u=-\sin (x) d(x)\)

\(\int \frac{\tan ^{3}(x)}{\cos ^{3}(x)} d x=\int \frac{\sin ^{2}(x) \sin (x)}{\cos ^{6}(x)} d x\)

\(\int \frac{\tan ^{3}(x)}{\cos ^{3}(x)} d x=-\int \frac{\sin ^{2}(x)^{(-)} \sin (x)}{\cos ^{6}(x)} d x\)

\(\sin ^{2}(x)=1-\cos ^{2}(x)\)

\(=-\int \frac{\left(1-\cos ^{2} (x)\right) d u}{u^{6}}\)

\(\int \frac{\tan ^{3}(x)}{\cos ^{3}(x)} d x=-\int \frac{\left(1-u^{2}\right)}{u^{6}} d u\)

\(=-\int \frac{1}{u^6} d u+\int \frac{u^{2}}{u^{6}} d u\)

\(=-\int u^{-6} d u+\int u^{-4} d u=-\frac{u^{-5}}{-5}+\frac{u^{-3}}{-3}+c\)

\(\int \frac{\tan ^{3}(x) d x}{\cos ^{3}(x)}=\frac{1}{5 u^{5}}+\frac{1}{-3 u^{3}}+c=\)

\(=\frac{1}{5 \cos ^{5} x}-\frac{1}{3 \cos ^{3}(x)}+C\)

Evaluate \(\int \sec ^{-1}(\sqrt{x}) d x\)

let \(t=\sqrt{x} \rightarrow t^{2}=x \rightarrow d x=2 t d t\)

\(\int \sec ^{-1}(\sqrt{x}) d x=\int 2 t \sec ^{-1}(t) d t\)

ILATE

\(\frac{d \sec ^{-1}(x)}{d x}=\frac{d x}{x \sqrt{x^{2}-1}}\)

\(u=\sec ^{-1}(t) \rightarrow d{u}=\frac{dt}{t \sqrt{t^{2}-1}}\)

\(v=t^{2} \quad \leftarrow d v=2td t\)

\(\int {2t} \sec ^{-1}(t) d t=u \cdot v-\int v d u=t^{2} \sec ^{-1}(t)-\int t^{2} \cdot \frac{d t}{t \sqrt{t^{2}-1}}\)

\(\int {2} t \sec ^{-1}(t) d t=t^{2} \sec ^{-1}(t)-\int \frac{t d t}{\sqrt{t^{2}-1}}+c\)

\(\int \frac{td t}{\sqrt{t^{2}-1}} \Rightarrow \text { let } u=t^{2}-1 \rightarrow d u=2 t d t\)

\(\frac{1}{2} \frac{\int {2} t d t}{\sqrt{t^{2}-1}}=\int \frac{d u}{\sqrt{u}}=\int \frac{d u}{u^{\frac{1}{2}}}=\int u^{-\frac{1}{2}} d u=2 u^{\frac{1}{2}}+c\)

\(\int {2}t\sec ^{-1}(t) d t=t^{2} \sec ^{-1}(t)-2\left(t^{2}-1\right)^{\frac{1}{2}}+c\)

\(\int \sec ^{-1}(\sqrt{x}) d x=x \sec ^{-1}(t)-2 \sqrt{x-1}+c\)

No comments yet

Join the conversation

Join Notatee Today!