- #1
Dethrone
- 717
- 0
$$\int \frac{x}{\sqrt{x^2-49}}dx$$
Method 1: (by inspection)
$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \,d\sqrt{x^2-49}=\sqrt{x^2-49}+C$$
Method 2: (Trig substitution)
$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \frac{7\sec(\theta)}{\sqrt{49\sec^2(\theta) -49}}d(7\sec(\theta))=7\int \frac{\sec^2(\theta)\tan(\theta)}{\left| \tan(\theta) \right|}$$
Let's assume $\tan(\theta)>0$
$$=7 \int \sec^2(\theta) \,d\theta=7\sqrt{x^2-49}+C$$
Why do the answers differ by a factor of $7$? I know it doesn't matter in integration, because you just want the difference between endpoints, i.e $F(b)-F(a)$=$7F(b)-7F(a)$, but it's the first time I've seen this.
Method 1: (by inspection)
$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \,d\sqrt{x^2-49}=\sqrt{x^2-49}+C$$
Method 2: (Trig substitution)
$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \frac{7\sec(\theta)}{\sqrt{49\sec^2(\theta) -49}}d(7\sec(\theta))=7\int \frac{\sec^2(\theta)\tan(\theta)}{\left| \tan(\theta) \right|}$$
Let's assume $\tan(\theta)>0$
$$=7 \int \sec^2(\theta) \,d\theta=7\sqrt{x^2-49}+C$$
Why do the answers differ by a factor of $7$? I know it doesn't matter in integration, because you just want the difference between endpoints, i.e $F(b)-F(a)$=$7F(b)-7F(a)$, but it's the first time I've seen this.